\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{Value of }\sin10^\circ+\sin20^\circ+\sin30^\circ+\ldots+\sin360^\circ\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad\text{(c) }2\qquad\text{(d) }1/2
\displaystyle \text{Answer:}
\displaystyle \text{We use }\sin(360^\circ-\theta)=-\sin\theta.
\displaystyle \sin10^\circ+\sin350^\circ=0,\quad\sin20^\circ+\sin340^\circ=0,\quad\ldots
\displaystyle \text{Also, }\sin180^\circ=0\text{ and }\sin360^\circ=0.
\displaystyle \therefore \sin10^\circ+\sin20^\circ+\sin30^\circ+\ldots+\sin360^\circ=0.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\sin A=\frac{3}{5},\ 0<A<\frac{\pi}{2}\text{ and }\cos B=-\frac{12}{13},
\displaystyle \pi<B<\frac{3\pi}{2},\text{ then value of }\sin(A-B)\text{ is}
\displaystyle \text{(a) }-\frac{13}{82}\qquad\text{(b) }-\frac{15}{65}\qquad\text{(c) }-\frac{13}{75}\qquad\text{(d) }-\frac{16}{65}
\displaystyle \text{Answer:}
\displaystyle \text{Since }0<A<\frac{\pi}{2},\text{ angle }A\text{ lies in the first quadrant.}
\displaystyle \cos A=\sqrt{1-\sin^2A}=\sqrt{1-\frac{9}{25}}=\frac{4}{5}.
\displaystyle \text{Since }\pi<B<\frac{3\pi}{2},\text{ angle }B\text{ lies in the third quadrant.}
\displaystyle \sin B=-\sqrt{1-\cos^2B}=-\sqrt{1-\frac{144}{169}}=-\frac{5}{13}.
\displaystyle \sin(A-B)=\sin A\cos B-\cos A\sin B.
\displaystyle =\frac{3}{5}\left(-\frac{12}{13}\right)-\frac{4}{5}\left(-\frac{5}{13}\right).
\displaystyle =-\frac{36}{65}+\frac{20}{65}=-\frac{16}{65}.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The large hand of a clock is }42\text{ cm long. How much distance does its}
\displaystyle \text{extremity move in }20\text{ minutes?}
\displaystyle \text{(a) }88\text{ cm}\qquad\text{(b) }80\text{ cm}\qquad\text{(c) }75\text{ cm}\qquad\text{(d) }77\text{ cm}
\displaystyle \text{Answer:}
\displaystyle \text{In }60\text{ minutes, the extremity of the minute hand completes one full circle.}
\displaystyle \text{In }20\text{ minutes, the angle swept }=\frac{20}{60}\times360^\circ=120^\circ.
\displaystyle \text{Distance moved}=\frac{120^\circ}{360^\circ}\times2\pi\times42.
\displaystyle =28\pi=28\times\frac{22}{7}=88\text{ cm}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The value of }\tan20^\circ+2\tan50^\circ-\tan70^\circ\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad\text{(c) }\tan50^\circ\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Using }\tan70^\circ=\cot20^\circ,\text{ we get}
\displaystyle \tan20^\circ-\tan70^\circ=\tan20^\circ-\cot20^\circ.
\displaystyle =\frac{\sin^2 20^\circ-\cos^2 20^\circ}{\sin20^\circ\cos20^\circ}.
\displaystyle =-\frac{2\cos40^\circ}{\sin40^\circ}=-2\cot40^\circ.
\displaystyle \text{Also, }\cot40^\circ=\tan50^\circ.
\displaystyle \therefore \tan20^\circ-\tan70^\circ=-2\tan50^\circ.
\displaystyle \therefore \tan20^\circ+2\tan50^\circ-\tan70^\circ=0.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\sin x+\cos x=\frac{1}{5}\text{ then }\tan2x\text{ is}
\displaystyle \text{(a) }\frac{25}{17}\qquad\text{(b) }\frac{7}{25}\qquad\text{(c) }\frac{25}{7}\qquad\text{(d) }\frac{24}{7}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin x+\cos x=\frac{1}{5}.
\displaystyle \text{Squaring both sides, we get}
\displaystyle \sin^2x+\cos^2x+2\sin x\cos x=\frac{1}{25}.
\displaystyle 1+\sin2x=\frac{1}{25}\implies\sin2x=-\frac{24}{25}.
\displaystyle \text{Also, }(\cos x-\sin x)^2=1-\sin2x=1+\frac{24}{25}=\frac{49}{25}.
\displaystyle \therefore \cos x-\sin x=\pm\frac{7}{5}.
\displaystyle \cos2x=(\cos x+\sin x)(\cos x-\sin x)=\pm\frac{7}{25}.
\displaystyle \therefore \tan2x=\frac{\sin2x}{\cos2x}=\pm\frac{24}{7}.
\displaystyle \text{Thus, without an additional condition on }x,\text{ the answer is not uniquely determined.}
\displaystyle \text{The intended answer appears to be }\frac{24}{7},\text{ i.e. option (d).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Consider the following statements.}
\displaystyle \textbf{I. }\cot x\text{ decreases from }0\text{ to }-\infty\text{ in first quadrant and increases from }0
\displaystyle \text{to }\infty\text{ in third quadrant.}
\displaystyle \textbf{II. }\sec x\text{ increases from }-\infty\text{ to }-1\text{ in second quadrant and decreases}
\displaystyle \text{from }\infty\text{ to }1\text{ in fourth quadrant.}
\displaystyle \textbf{III. }\mathrm{cosec}\,x\text{ increases from }1\text{ to }\infty\text{ in second quadrant and decreases}
\displaystyle \text{from }-1\text{ to }-\infty\text{ in fourth quadrant.}
\displaystyle \text{Choose the correct option.}
\displaystyle \text{(a) I is incorrect}\qquad\text{(b) II is incorrect}
\displaystyle \text{(c) III is incorrect}\qquad\text{(d) IV is incorrect}
\displaystyle \text{Answer:}
\displaystyle \text{In the first quadrant, }\cot x\text{ decreases from }\infty\text{ to }0.
\displaystyle \text{In the third quadrant also, }\cot x\text{ decreases from }\infty\text{ to }0.
\displaystyle \text{Hence, statement I is incorrect.}
\displaystyle \text{Statement II correctly describes the variation of }\sec x\text{ in the stated quadrants.}
\displaystyle \text{Statement III correctly describes the variation of }\mathrm{cosec}\,x\text{ in the stated quadrants.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 7: }1+\cos2x+\cos4x+\cos6x=
\displaystyle \text{(a) }2\cos x\cos2x\cos3x\qquad\text{(b) }4\sin x\cos2x\cos3x
\displaystyle \text{(c) }4\cos x\cos2x\cos3x\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle 1+\cos6x=2\cos^2 3x.
\displaystyle \cos2x+\cos4x=2\cos3x\cos x.
\displaystyle \therefore 1+\cos2x+\cos4x+\cos6x=2\cos3x(\cos3x+\cos x).
\displaystyle \text{Using }\cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2},\text{ we get}
\displaystyle \cos3x+\cos x=2\cos2x\cos x.
\displaystyle \therefore 1+\cos2x+\cos4x+\cos6x=4\cos x\cos2x\cos3x.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The value of }\sin\left(-\frac{11\pi}{3}\right)\text{ is }\frac{\sqrt{3}}{m}.\text{ Value of `m' is}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \text{Using the periodicity of sine, }\sin(\theta+4\pi)=\sin\theta.
\displaystyle \sin\left(-\frac{11\pi}{3}\right)=\sin\left(-\frac{11\pi}{3}+4\pi\right).
\displaystyle =\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}.
\displaystyle \text{Comparing with }\frac{\sqrt{3}}{m},\text{ we get }m=2.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The number of solutions }\sin x=\cos x\text{ for all }x\in[-\pi,\pi]\text{ from graph will be}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \sin x=\cos x\implies\tan x=1.
\displaystyle \therefore x=\frac{\pi}{4}+n\pi,\quad n\in Z.
\displaystyle \text{In the interval }[-\pi,\pi],\text{ the solutions are }x=-\frac{3\pi}{4},\frac{\pi}{4}.
\displaystyle \therefore \text{The number of solutions is }2.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The value of expression }\sin A+\cos A\text{ lies between}
\displaystyle \text{(a) }-\sqrt{2}\text{ and }\sqrt{2}\text{ both inclusive}\qquad\text{(b) }0\text{ and }2\text{ both inclusive}
\displaystyle \text{(c) }-2\text{ and }2\text{ both inclusive}\qquad\text{(d) }0\text{ and }\sqrt{2}\text{ both inclusive}
\displaystyle \text{Answer:}
\displaystyle \sin A+\cos A=\sqrt{2}\sin\left(A+\frac{\pi}{4}\right).
\displaystyle \text{Since }-1\leq\sin\left(A+\frac{\pi}{4}\right)\leq1,\text{ we get}
\displaystyle -\sqrt{2}\leq\sin A+\cos A\leq\sqrt{2}.
\displaystyle \therefore \sin A+\cos A\text{ lies between }-\sqrt{2}\text{ and }\sqrt{2}\text{ both inclusive.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\sin\theta+\cos\theta=1,\text{ then }\sin\theta\cdot\cos\theta=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }1/2
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin\theta+\cos\theta=1.
\displaystyle \text{Squaring both sides, we get}
\displaystyle \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=1.
\displaystyle 1+2\sin\theta\cos\theta=1.
\displaystyle \therefore \sin\theta\cos\theta=0.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The degree measure of }\frac{11}{16}\text{ radian is}
\displaystyle \text{(a) }39^\circ22'30''\qquad\text{(b) }39^\circ22.3'\qquad\text{(c) }39^\circ30'22''\qquad\text{(d) }39^\circ
\displaystyle \text{Answer:}
\displaystyle \text{We know that }1\text{ radian}=\frac{180^\circ}{\pi}.
\displaystyle \therefore \frac{11}{16}\text{ radian}=\frac{11}{16}\times\frac{180^\circ}{\pi}.
\displaystyle \text{Taking }\pi=\frac{22}{7},\text{ we get}
\displaystyle \frac{11}{16}\times\frac{180\times7}{22}=39.375^\circ.
\displaystyle 0.375^\circ=0.375\times60'=22.5'=22'30''.
\displaystyle \therefore \frac{11}{16}\text{ radian}=39^\circ22'30''.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If in two circles, arcs of the same length subtend angles }45^\circ\text{ and }60^\circ
\displaystyle \text{at Centre, find the ratio of their radii.}
\displaystyle \text{(a) }2:3\qquad\text{(b) }2:5\qquad\text{(c) }3:4\qquad\text{(d) }4:3
\displaystyle \text{Answer:}
\displaystyle \text{For an arc of length }l,\text{ we have }l=r\theta.
\displaystyle \text{Since the lengths of the arcs are equal, }r_1\theta_1=r_2\theta_2.
\displaystyle \therefore \frac{r_1}{r_2}=\frac{\theta_2}{\theta_1}=\frac{60^\circ}{45^\circ}=\frac{4}{3}.
\displaystyle \therefore r_1:r_2=4:3.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Consider the statements given below:}
\displaystyle \text{I. }\sin x\text{ is positive in first and second quadrants.}
\displaystyle \text{II. }\mathrm{cosec}\,x\text{ is negative in third and fourth quadrants.}
\displaystyle \text{III. }\tan x\text{ and }\cot x\text{ are negative in second and fourth quadrants.}
\displaystyle \text{IV. }\cos x\text{ and }\sec x\text{ are positive in first and fourth quadrants. Choose}
\displaystyle \text{the correct option.}
\displaystyle \text{(a) All are correct}\qquad\text{(b) Only I and IV are correct}
\displaystyle \text{(c) Only III and IV are correct}\qquad\text{(d) None is correct}
\displaystyle \text{Answer:}
\displaystyle \text{In the first and second quadrants, }\sin x>0,\text{ so statement I is correct.}
\displaystyle \text{Since }\mathrm{cosec}\,x=\frac{1}{\sin x},\text{ it is negative in third and fourth quadrants.}
\displaystyle \text{Hence, statement II is correct.}
\displaystyle \tan x\text{ and }\cot x\text{ are negative in second and fourth quadrants.}
\displaystyle \text{Hence, statement III is correct.}
\displaystyle \cos x\text{ and }\sec x\text{ are positive in first and fourth quadrants.}
\displaystyle \text{Hence, statement IV is correct.}
\displaystyle \therefore \text{All the statements are correct.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The value of }\sin(\pi-\theta)\sin(\pi+\theta)\mathrm{cosec}^2\theta=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }-1\qquad\text{(d) }1/2
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\sin(\pi-\theta)=\sin\theta\text{ and }\sin(\pi+\theta)=-\sin\theta.
\displaystyle \therefore \sin(\pi-\theta)\sin(\pi+\theta)\mathrm{cosec}^2\theta
\displaystyle =(\sin\theta)(-\sin\theta)\frac{1}{\sin^2\theta}=-1.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\tan A=1/2\text{ and }\tan B=1/3,\text{ then the value of }A+B=
\displaystyle \text{(a) }45^\circ\qquad\text{(b) }60^\circ\qquad\text{(c) }30^\circ\qquad\text{(d) }0^\circ
\displaystyle \text{Answer:}
\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.
\displaystyle =\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times\frac{1}{3}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1.
\displaystyle \therefore \tan(A+B)=1.
\displaystyle \therefore A+B=45^\circ.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The value of }\tan3A-\tan2A-\tan A\text{ is equal to}
\displaystyle \text{(a) }\tan3A\tan2A\tan A
\displaystyle \text{(b) }-\tan3A\tan2A\tan A
\displaystyle \text{(c) }\tan A\tan2A-\tan2A\tan3A-\tan3A\tan A
\displaystyle \text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Since }3A=2A+A,\text{ we have }\tan3A=\tan(2A+A).
\displaystyle \tan3A=\frac{\tan2A+\tan A}{1-\tan2A\tan A}.
\displaystyle \tan3A(1-\tan2A\tan A)=\tan2A+\tan A.
\displaystyle \tan3A-\tan2A-\tan A=\tan3A\tan2A\tan A.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The value of }\tan75^\circ-\cot75^\circ\text{ is equal to}
\displaystyle \text{(a) }2\sqrt{3}\qquad\text{(b) }2+\sqrt{3}\qquad\text{(c) }2-\sqrt{3}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \tan75^\circ=\tan(45^\circ+30^\circ).
\displaystyle =\frac{\tan45^\circ+\tan30^\circ}{1-\tan45^\circ\tan30^\circ}.
\displaystyle =\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}=2+\sqrt{3}.
\displaystyle \therefore \cot75^\circ=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}.
\displaystyle \therefore \tan75^\circ-\cot75^\circ=(2+\sqrt{3})-(2-\sqrt{3})=2\sqrt{3}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The value of }\frac{1}{\sin10^\circ}-\frac{\sqrt{3}}{\cos10^\circ}=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }-1\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \frac{1}{\sin10^\circ}-\frac{\sqrt{3}}{\cos10^\circ}
\displaystyle =\frac{\cos10^\circ-\sqrt{3}\sin10^\circ}{\sin10^\circ\cos10^\circ}.
\displaystyle \cos10^\circ-\sqrt{3}\sin10^\circ=2\cos(10^\circ+60^\circ)=2\cos70^\circ.
\displaystyle =2\sin20^\circ.
\displaystyle \text{Also, }\sin10^\circ\cos10^\circ=\frac{1}{2}\sin20^\circ.
\displaystyle \therefore \frac{2\sin20^\circ}{\frac{1}{2}\sin20^\circ}=4.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }\tan A=b/a,\text{ then value of }a\cos2A+b\sin2A=
\displaystyle \text{(a) }-b\qquad\text{(b) }a\qquad\text{(c) }-a\qquad\text{(d) }b
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\tan A=\frac{b}{a}.
\displaystyle \cos2A=\frac{1-\tan^2A}{1+\tan^2A}=\frac{a^2-b^2}{a^2+b^2}.
\displaystyle \sin2A=\frac{2\tan A}{1+\tan^2A}=\frac{2ab}{a^2+b^2}.
\displaystyle \therefore a\cos2A+b\sin2A
\displaystyle =a\left(\frac{a^2-b^2}{a^2+b^2}\right)+b\left(\frac{2ab}{a^2+b^2}\right).
\displaystyle =\frac{a^3-ab^2+2ab^2}{a^2+b^2}=\frac{a(a^2+b^2)}{a^2+b^2}=a.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

 

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED}

\displaystyle \textbf{Question 21: }\text{CASE STUDY -1 :- Trigonometry is the combination of 2 words - `Trigonon'}
\displaystyle \text{means Triangle and `metron' means measure. It is a branch of geometry that studies}
\displaystyle \text{relationship between lengths and angles of a triangle. Degree and radian units of}
\displaystyle \text{measurement of angles are used, also called Indian system of measurement of triangles.}
\displaystyle \text{In this system }\pi\text{ radian}=180^\circ;\ 1^\circ=60\text{ minute and }1\text{ minute}=60\text{ seconds;}
\displaystyle 1\text{ rt angle}=90^\circ.\text{ The length of arc }l\text{ is given by }\theta=lr.\text{ On the basis of above}
\displaystyle \text{information answer the following questions:}

\displaystyle \text{(i) }11/36\text{ radians into degree minutes and seconds}
\displaystyle \text{(a) }17^\circ14'30''\qquad\text{(b) }17^\circ14'\qquad\text{(c) }17^\circ30''\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \frac{11}{36}\text{ radian}=\frac{11}{36}\times\frac{180^\circ}{\pi}.
\displaystyle \text{Using }\pi=\frac{22}{7},\text{ we get}
\displaystyle \frac{11}{36}\times\frac{180\times7}{22}=17.5^\circ.
\displaystyle 0.5^\circ=30'.
\displaystyle \therefore \frac{11}{36}\text{ radian}=17^\circ30'.
\displaystyle \text{Hence, none of the given options matches exactly.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{(ii) }\frac{7\pi}{18}\text{ into degrees will be}
\displaystyle \text{(a) }60^\circ\qquad\text{(b) }70^\circ\qquad\text{(c) }100^\circ\qquad\text{(d) }80^\circ
\displaystyle \text{Answer:}
\displaystyle \frac{7\pi}{18}\text{ radian}=\frac{7\pi}{18}\times\frac{180^\circ}{\pi}.
\displaystyle =70^\circ.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{(iii) Find the length of arc made by minute's hand of a clock in 5 minutes having radius }7\text{ cm.}
\displaystyle \text{(a) }\frac{7\pi}{5}\qquad\text{(b) }\frac{7\pi}{6}\qquad\text{(c) }\frac{7\pi}{3}\qquad\text{(d) }\frac{7\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \text{In }60\text{ minutes, the minute hand completes an angle of }2\pi\text{ radians.}
\displaystyle \text{In }5\text{ minutes, }\theta=\frac{5}{60}\times2\pi=\frac{\pi}{6}.
\displaystyle \text{Using }l=r\theta,\text{ we get}
\displaystyle l=7\times\frac{\pi}{6}=\frac{7\pi}{6}\text{ cm}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{(iv) If the arcs of the same length in two circles subtend angles }65\text{ and }80\text{ at the centre,}
\displaystyle \text{then the ratio of their radii.}
\displaystyle \text{(a) }13:16\qquad\text{(b) }3:16\qquad\text{(c) }16:13\qquad\text{(d) }5:16
\displaystyle \text{Answer:}
\displaystyle \text{For equal arc lengths, }r_1\theta_1=r_2\theta_2.
\displaystyle \therefore \frac{r_1}{r_2}=\frac{\theta_2}{\theta_1}=\frac{80}{65}=\frac{16}{13}.
\displaystyle \therefore r_1:r_2=16:13.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{CASE STUDY-2-Domain and range of the following trigonometric}
\displaystyle \text{functions are given below:-}
\displaystyle \text{Based on the following information, answer the following questions:-}
\displaystyle \text{Domain and Range of trigonometric functions}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Function} & \text{Domain} & \text{Range} \\ \hline \sin x & R & [-1,1] \\ \hline \cos x & R & [-1,1] \\ \hline \tan x & R-\left\{(2n+1)\frac{\pi}{2}:n\in N\right\} & R \\ \hline \mathrm{cosec}\,x & R-\{n\pi:n\in N\} & R-(-1,1) \\ \hline \sec x & R-\left\{(2n+1)\frac{\pi}{2}:n\in N\right\} & R-(-1,1) \\ \hline \cot x & R-\{n\pi:n\in N\} & R \\ \hline \end{array}

\displaystyle \text{(i) Domain of }f(x)=\frac{1}{\sqrt{1+\cos x}}\text{ is}
\displaystyle \text{(a) }R\qquad\text{(b) }R-(2n+1)\pi\qquad\text{(c) }R-2n\pi\qquad\text{(d) }R-n\pi
\displaystyle \text{Answer:}
\displaystyle \text{For the function to be defined, the denominator must be non-zero.}
\displaystyle 1+\cos x>0.
\displaystyle 1+\cos x=0\implies\cos x=-1.
\displaystyle \therefore x=(2n+1)\pi,\quad n\in Z.
\displaystyle \therefore \text{Domain}=R-\{(2n+1)\pi:n\in Z\}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{(ii) Domain of }f(x)=\frac{1}{\sin x+\cos x}\text{ is}
\displaystyle \text{(a) }R-(4n-1)\frac{\pi}{4}\qquad\text{(b) }R-(4n+1)\frac{\pi}{4}
\displaystyle \text{(c) }R-\frac{n\pi}{4}\qquad\text{(d) }R-n\pi
\displaystyle \text{Answer:}
\displaystyle \text{The denominator must not be equal to zero.}
\displaystyle \sin x+\cos x\neq0.
\displaystyle \sin x+\cos x=0\implies\tan x=-1.
\displaystyle \therefore x=-\frac{\pi}{4}+n\pi=\frac{(4n-1)\pi}{4},\quad n\in Z.
\displaystyle \therefore \text{Domain}=R-\left\{\frac{(4n-1)\pi}{4}:n\in Z\right\}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(iii) Range of }f(x)=\sin2x
\displaystyle \text{(a) }[-1,1]\qquad\text{(b) }(-1,1)\qquad\text{(c) }[0,1]\qquad\text{(d) }(-1,0)
\displaystyle \text{Answer:}
\displaystyle \text{For every real value of }x,\ -1\leq\sin2x\leq1.
\displaystyle \therefore \text{Range of }f=[-1,1].
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(iv) Range of }f(x)=\sin x+\cos x
\displaystyle \text{(a) }(-\sqrt{2},\sqrt{2})\qquad\text{(b) }[-\sqrt{2},\sqrt{2}]
\displaystyle \text{(c) }(-\sqrt{2},0)\qquad\text{(d) }(0,\sqrt{2})
\displaystyle \text{Answer:}
\displaystyle \sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right).
\displaystyle \text{Since }-1\leq\sin\left(x+\frac{\pi}{4}\right)\leq1,\text{ we get}
\displaystyle -\sqrt{2}\leq\sin x+\cos x\leq\sqrt{2}.
\displaystyle \therefore \text{Range of }f=[-\sqrt{2},\sqrt{2}].
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{CASE STUDY -3: Sine and cosine functions can be used to model many}
\displaystyle \text{real life scenarios-radio waves, tides, musical tones, electrical signals. Following are shown}
\displaystyle \text{the graphs of sine and cosine curves. Looking to these graphs, answer the following questions:-}

\displaystyle \text{(i) If we draw line }y=1/2,\text{ at how many points the graph of sine and this line intersects}
\displaystyle \text{between interval }[-\pi,\pi]
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \text{We have to find the number of solutions of }\sin x=\frac{1}{2}\text{ in }[-\pi,\pi].
\displaystyle \sin x=\frac{1}{2}\text{ at }x=\frac{\pi}{6}\text{ and }x=\frac{5\pi}{6}\text{ in the given interval.}
\displaystyle \therefore \text{The line intersects the sine curve at }2\text{ points.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \text{(ii) If both sine and cosine curve are drawn on the same graph and on the same interval}
\displaystyle [-2\pi,2\pi]
\displaystyle \text{(a) }4\qquad\text{(b) }5\qquad\text{(c) }6\qquad\text{(d) }8
\displaystyle \text{Answer:}
\displaystyle \text{The sine and cosine curves intersect when }\sin x=\cos x.
\displaystyle \tan x=1\implies x=\frac{\pi}{4}+n\pi,\quad n\in Z.
\displaystyle \text{In }[-2\pi,2\pi],\text{ there are }4\text{ such values of }x.
\displaystyle \therefore \text{The sine and cosine curves intersect at }4\text{ points.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(iii) The graph of cosine cuts on X axis between }(0,2\pi)
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \text{The cosine curve cuts the X-axis where }\cos x=0.
\displaystyle \text{In }(0,2\pi),\text{ we have }x=\frac{\pi}{2}\text{ and }x=\frac{3\pi}{2}.
\displaystyle \therefore \text{The cosine curve cuts the X-axis at }2\text{ points.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \text{(iv) At how many points will be }\sin x=\sqrt{2}\text{ in interval }[-\pi,\pi]
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \text{For every real value of }x,\text{ we have }-1\leq\sin x\leq1.
\displaystyle \text{Since }\sqrt{2}>1,\text{ the equation }\sin x=\sqrt{2}\text{ has no real solution.}
\displaystyle \therefore \text{There are }0\text{ such points in the interval }[-\pi,\pi].
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{CASE STUDY-5 }\text{It can be obtained that }\sin18^\circ=\frac{\sqrt{5}-1}{4}\text{ and }\cos36^\circ=
\displaystyle \frac{\sqrt{5}+1}{4}.\text{ Using these values, answer the following:}

\displaystyle \text{(i) }\sin47^\circ+\sin18^\circ-\sin11^\circ-\sin25^\circ\text{ is equal to}
\displaystyle \text{(a) }\sin36^\circ\qquad\text{(b) }\cos36^\circ\qquad\text{(c) }\sin7^\circ\qquad\text{(d) }\cos7^\circ
\displaystyle \text{Answer:}
\displaystyle \sin47^\circ+\sin18^\circ-\sin11^\circ-\sin25^\circ\approx0.42694.
\displaystyle \sin36^\circ\approx0.58779,\quad\cos36^\circ\approx0.80902.
\displaystyle \sin7^\circ\approx0.12187,\quad\cos7^\circ\approx0.99255.
\displaystyle \text{Hence, the given expression does not match any of the four options.}
\displaystyle \text{Therefore, there appears to be an error in part (i) of the question or its options.}
\displaystyle \\

\displaystyle \text{(ii) }\cos12^\circ+\cos84^\circ+\cos156^\circ-\cos132^\circ\text{ is equal to}
\displaystyle \text{(a) }-1\qquad\text{(b) }-\frac{1}{2}\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \cos12^\circ+\cos84^\circ+\cos156^\circ-\cos132^\circ\approx0.83826.
\displaystyle \text{This value is neither }-1,\ -\frac{1}{2},\ \frac{1}{2}\text{ nor }1.
\displaystyle \text{Hence, there appears to be an error in part (ii) of the question or its options.}
\displaystyle \\

\displaystyle \text{(iii) }\sin6^\circ-\sin66^\circ+\sin78^\circ-\sin42^\circ\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }-\frac{1}{2}\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \sin6^\circ-\sin66^\circ=2\cos36^\circ\sin(-30^\circ)=-\cos36^\circ.
\displaystyle \sin78^\circ-\sin42^\circ=2\cos60^\circ\sin18^\circ=\sin18^\circ.
\displaystyle \therefore \sin6^\circ-\sin66^\circ+\sin78^\circ-\sin42^\circ
\displaystyle =\sin18^\circ-\cos36^\circ.
\displaystyle =\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}=-\frac{1}{2}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \text{(iv) }\tan81^\circ-\tan63^\circ-\tan27^\circ+\tan9^\circ\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \tan81^\circ+\tan9^\circ=\frac{\sin90^\circ}{\cos81^\circ\cos9^\circ}.
\displaystyle =\frac{1}{\sin9^\circ\cos9^\circ}=\frac{2}{\sin18^\circ}.
\displaystyle \tan63^\circ+\tan27^\circ=\frac{\sin90^\circ}{\cos63^\circ\cos27^\circ}.
\displaystyle =\frac{1}{\sin27^\circ\cos27^\circ}=\frac{2}{\sin54^\circ}.
\displaystyle \therefore \tan81^\circ-\tan63^\circ-\tan27^\circ+\tan9^\circ
\displaystyle =\frac{2}{\sin18^\circ}-\frac{2}{\sin54^\circ}.
\displaystyle \text{Using }\sin18^\circ=\frac{\sqrt{5}-1}{4}\text{ and }\sin54^\circ=\cos36^\circ=\frac{\sqrt{5}+1}{4},
\displaystyle =\frac{8}{\sqrt{5}-1}-\frac{8}{\sqrt{5}+1}.
\displaystyle =2(\sqrt{5}+1)-2(\sqrt{5}-1)=4.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 25: } \text{CASE STUDY-5 } \text{Given }\cos x=-\frac{4}{5}\text{ and }\sin y=\frac{5}{13},\ x\text{ \& }y\text{ both lie in second}
\displaystyle \text{quadrant. Based on the following questions, answer the following:-}

\displaystyle \text{(i) The value of }\sin(x+y)=
\displaystyle \text{(a) }-\frac{56}{65}\qquad\text{(b) }-\frac{33}{65}\qquad\text{(c) }-\frac{16}{65}\qquad\text{(d) }\frac{63}{65}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\sin x>0.
\displaystyle \sin x=\sqrt{1-\cos^2x}=\sqrt{1-\frac{16}{25}}=\frac{3}{5}.
\displaystyle \text{Since }y\text{ lies in the second quadrant, }\cos y<0.
\displaystyle \cos y=-\sqrt{1-\sin^2y}=-\sqrt{1-\frac{25}{169}}=-\frac{12}{13}.
\displaystyle \sin(x+y)=\sin x\cos y+\cos x\sin y.
\displaystyle =\frac{3}{5}\left(-\frac{12}{13}\right)+\left(-\frac{4}{5}\right)\frac{5}{13}=-\frac{56}{65}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{(ii) The value of }\cos(x+y)=
\displaystyle \text{(a) }-\frac{56}{65}\qquad\text{(b) }-\frac{33}{65}\qquad\text{(c) }-\frac{16}{65}\qquad\text{(d) }\frac{33}{65}
\displaystyle \text{Answer:}
\displaystyle \cos(x+y)=\cos x\cos y-\sin x\sin y.
\displaystyle =\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)-\frac{3}{5}\cdot\frac{5}{13}.
\displaystyle =\frac{48}{65}-\frac{15}{65}=\frac{33}{65}.
\displaystyle \therefore \cos(x+y)=\frac{33}{65}.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{(iii) The value of }\sin(x-y)=
\displaystyle \text{(a) }-\frac{56}{65}\qquad\text{(b) }-\frac{33}{65}\qquad\text{(c) }-\frac{16}{65}\qquad\text{(d) }\frac{63}{65}
\displaystyle \text{Answer:}
\displaystyle \sin(x-y)=\sin x\cos y-\cos x\sin y.
\displaystyle =\frac{3}{5}\left(-\frac{12}{13}\right)-\left(-\frac{4}{5}\right)\frac{5}{13}.
\displaystyle =-\frac{36}{65}+\frac{20}{65}=-\frac{16}{65}.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \text{(iv) The value of }\cos(x-y)=
\displaystyle \text{(a) }-\frac{56}{65}\qquad\text{(b) }-\frac{33}{65}\qquad\text{(c) }-\frac{16}{65}\qquad\text{(d) }\frac{63}{65}
\displaystyle \text{Answer:}
\displaystyle \cos(x-y)=\cos x\cos y+\sin x\sin y.
\displaystyle =\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\frac{3}{5}\cdot\frac{5}{13}.
\displaystyle =\frac{48}{65}+\frac{15}{65}=\frac{63}{65}.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{ASSERTION REASONING }

\displaystyle \text{Each of these questions contains two statements, Assertion and Reason. Each of these}
\displaystyle \text{questions also has four alternative choices, only one of which is the correct answer. You have}
\displaystyle \text{to select one of the codes (a), (b), (c) and (d) given below.}

\displaystyle \textbf{Question 26: }\text{Assertion: The ratio of the radii of two circles at the centres of which}
\displaystyle \text{two equal arcs subtend angles of }30^\circ\text{ and }70^\circ\text{ is }21:10.
\displaystyle \text{Reason: Number of radians in an angle subtended at the centre of a circle by an arc is}
\displaystyle \text{equal to the ratio of the length of the arc to the radius of the circle.}
\displaystyle \text{(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.}
\displaystyle \text{(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.}
\displaystyle \text{(c) Assertion is correct, reason is incorrect.}
\displaystyle \text{(d) Assertion is incorrect, reason is correct.}
\displaystyle \text{Answer:}
\displaystyle \text{For an arc of length }l,\text{ the angle subtended at the centre is }\theta=\frac{l}{r}.
\displaystyle \text{Since the two arcs have equal lengths, }r_1\theta_1=r_2\theta_2.
\displaystyle \therefore \frac{r_1}{r_2}=\frac{\theta_2}{\theta_1}=\frac{70^\circ}{30^\circ}=\frac{7}{3}.
\displaystyle \therefore r_1:r_2=7:3,\text{ not }21:10.
\displaystyle \text{Hence, Assertion is incorrect.}
\displaystyle \text{Reason correctly states that }\theta=\frac{l}{r}\text{ when }\theta\text{ is measured in radians.}
\displaystyle \text{Hence, Reason is correct.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion: }\mathrm{cosec}\,x\text{ is negative in third and fourth quadrants.}
\displaystyle \text{Reason: }\cot x\text{ decreases from }0\text{ to }-\infty\text{ in first quadrant and increases from }0
\displaystyle \text{to }\infty\text{ in third quadrant.}
\displaystyle \text{(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.}
\displaystyle \text{(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.}
\displaystyle \text{(c) Assertion is correct, reason is incorrect.}
\displaystyle \text{(d) Assertion is incorrect, reason is correct.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }\sin x<0\text{ in the third and fourth quadrants, }\mathrm{cosec}\,x=\frac{1}{\sin x}<0.
\displaystyle \text{Hence, Assertion is correct.}
\displaystyle \text{In the first quadrant, }\cot x\text{ decreases from }\infty\text{ to }0.
\displaystyle \text{In the third quadrant also, }\cot x\text{ decreases from }\infty\text{ to }0.
\displaystyle \text{Hence, Reason is incorrect.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion: }
\displaystyle \cos^3\alpha+\cos^3\left(\alpha+\frac{2\pi}{3}\right)+\cos^3\left(\alpha+\frac{4\pi}{3}\right)=3\cos\alpha\cos\left(\alpha+\frac{2\pi}{3}\right)   \cos\left(\alpha+\frac{4\pi}{3}\right)
\displaystyle \text{Reason: If }a+b+c=0\text{ then }a^3+b^3+c^3=3abc.
\displaystyle \text{(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.}
\displaystyle \text{(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.}
\displaystyle \text{(c) Assertion is correct, reason is incorrect.}
\displaystyle \text{(d) Assertion is incorrect, reason is correct.}
\displaystyle \text{Answer:}
\displaystyle \text{We know that }\cos\alpha+\cos\left(\alpha+\frac{2\pi}{3}\right)+\cos\left(\alpha+\frac{4\pi}{3}\right)=0.
\displaystyle \text{Let }a=\cos\alpha,\quad b=\cos\left(\alpha+\frac{2\pi}{3}\right),\quad c=\cos\left(\alpha+\frac{4\pi}{3}\right).
\displaystyle \text{Then }a+b+c=0.
\displaystyle \text{Using }a^3+b^3+c^3=3abc,\text{ we get}
\displaystyle \cos^3\alpha+\cos^3\left(\alpha+\frac{2\pi}{3}\right)+\cos^3\left(\alpha+\frac{4\pi}{3}\right)
\displaystyle =3\cos\alpha\cos\left(\alpha+\frac{2\pi}{3}\right)\cos\left(\alpha+\frac{4\pi}{3}\right).
\displaystyle \text{Hence, Assertion is correct and Reason is also correct.}
\displaystyle \text{Reason gives the correct explanation of the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Assertion: }\tan5A-\tan3A-\tan2A=\tan5A\tan3A\tan2A
\displaystyle \text{Reason: }x=y+z\Rightarrow\tan x-\tan y-\tan z=\tan x\tan y\tan z.
\displaystyle \text{(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.}
\displaystyle \text{(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.}
\displaystyle \text{(c) Assertion is correct, reason is incorrect.}
\displaystyle \text{(d) Assertion is incorrect, reason is correct.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }5A=3A+2A,\text{ we use the tangent addition formula.}
\displaystyle \tan5A=\frac{\tan3A+\tan2A}{1-\tan3A\tan2A}.
\displaystyle \tan5A(1-\tan3A\tan2A)=\tan3A+\tan2A.
\displaystyle \therefore \tan5A-\tan3A-\tan2A=\tan5A\tan3A\tan2A.
\displaystyle \text{Hence, Assertion is correct.}
\displaystyle \text{Reason states the same identity for }x=y+z\text{ and correctly explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Assertion: The maximum value of }\sin x+\cos x\text{ is }2.
\displaystyle \text{Reason: The maximum value of }\sin x\text{ is }1\text{ and maximum value of }\cos x\text{ is }1.
\displaystyle \text{(a) Assertion is correct, reason is correct; reason is a correct explanation for assertion.}
\displaystyle \text{(b) Assertion is correct, reason is correct; reason is not a correct explanation for assertion.}
\displaystyle \text{(c) Assertion is correct, reason is incorrect.}
\displaystyle \text{(d) Assertion is incorrect, reason is correct.}
\displaystyle \text{Answer:}
\displaystyle \sin x+\cos x=\sqrt{2}\sin\left(x+\frac{\pi}{4}\right).
\displaystyle \text{Since the maximum value of sine is }1,\text{ the maximum value of }\sin x+\cos x\text{ is }\sqrt{2}.
\displaystyle \text{Hence, Assertion is incorrect.}
\displaystyle \text{The maximum value of }\sin x\text{ is }1\text{ and the maximum value of }\cos x\text{ is also }1.
\displaystyle \text{Hence, Reason is correct.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \text{VERY SHORT}

\displaystyle \textbf{Question 31: }\text{If }\sin(A+B)=m\text{ and }\sin(A-B)=n\text{ then find } \\ \cos^2B-\cos^2A
\displaystyle \text{Answer:}
\displaystyle mn=\sin(A+B)\sin(A-B).
\displaystyle =\sin^2A-\sin^2B.
\displaystyle =(1-\cos^2A)-(1-\cos^2B)=\cos^2B-\cos^2A.
\displaystyle \therefore \cos^2B-\cos^2A=mn.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{If the arcs of the same lengths in two circles subtend angles }65^\circ\text{ and}
\displaystyle 110^\circ\text{ at the centre, find the ratio of their radii}
\displaystyle \text{Answer:}
\displaystyle \text{For equal arc lengths, }r_1\theta_1=r_2\theta_2.
\displaystyle \therefore \frac{r_1}{r_2}=\frac{\theta_2}{\theta_1}=\frac{110}{65}=\frac{22}{13}.
\displaystyle \therefore r_1:r_2=22:13.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If }\tan(A-B)=1,\ \sec(A+B)=\frac{2}{\sqrt{3}}\text{ the smallest positive value of }B
\displaystyle \text{is........................}
\displaystyle \text{Answer:}
\displaystyle \tan(A-B)=1\implies A-B=-135^\circ.
\displaystyle \sec(A+B)=\frac{2}{\sqrt{3}}\implies\cos(A+B)=\frac{\sqrt{3}}{2}.
\displaystyle \text{For the smallest positive }B,\text{ take }A+B=-30^\circ.
\displaystyle 2B=(A+B)-(A-B)=-30^\circ-(-135^\circ)=105^\circ.
\displaystyle \therefore B=52.5^\circ=52^\circ30'.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{Find the value of }\cos^2 75^\circ-\sin^2 15^\circ
\displaystyle \text{Answer:}
\displaystyle \cos75^\circ=\sin15^\circ.
\displaystyle \therefore \cos^2 75^\circ-\sin^2 15^\circ=0.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Evaluate : }\sqrt{3}\,\mathrm{cosec}\,20^\circ-\sec20^\circ
\displaystyle \text{Answer:}
\displaystyle \sqrt{3}\,\mathrm{cosec}\,20^\circ-\sec20^\circ
\displaystyle =\frac{\sqrt{3}}{\sin20^\circ}-\frac{1}{\cos20^\circ}.
\displaystyle =\frac{\sqrt{3}\cos20^\circ-\sin20^\circ}{\sin20^\circ\cos20^\circ}.
\displaystyle =\frac{2\sin(60^\circ-20^\circ)}{\frac{1}{2}\sin40^\circ}.
\displaystyle =\frac{2\sin40^\circ}{\frac{1}{2}\sin40^\circ}=4.
\displaystyle \therefore \text{The required value is }4.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Find the value of }\cos\left(\frac{3\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}+x\right)
\displaystyle \text{Answer:}
\displaystyle \text{Using }\cos(A-B)-\cos(A+B)=2\sin A\sin B,
\displaystyle \cos\left(\frac{3\pi}{4}-x\right)-\cos\left(\frac{3\pi}{4}+x\right)
\displaystyle =2\sin\frac{3\pi}{4}\sin x=2\left(\frac{1}{\sqrt{2}}\right)\sin x.
\displaystyle \therefore \text{The required value is }\sqrt{2}\sin x.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Evaluate: }\sin\frac{\pi}{10}+\sin\frac{13\pi}{10}
\displaystyle \text{Answer:}
\displaystyle \sin\frac{13\pi}{10}=\sin\left(\pi+\frac{3\pi}{10}\right)=-\sin\frac{3\pi}{10}.
\displaystyle \therefore \sin\frac{\pi}{10}+\sin\frac{13\pi}{10}=\sin18^\circ-\sin54^\circ.
\displaystyle =\frac{\sqrt{5}-1}{4}-\frac{\sqrt{5}+1}{4}=-\frac{1}{2}.
\displaystyle \therefore \text{The required value is }-\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Evaluate : }2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}
\displaystyle \text{Answer:}
\displaystyle \text{Using }2\cos A\cos B=\cos(A+B)+\cos(A-B),
\displaystyle 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}=\cos\frac{10\pi}{13}+\cos\frac{8\pi}{13}.
\displaystyle \text{Also, }\cos\frac{10\pi}{13}=-\cos\frac{3\pi}{13}\text{ and }\cos\frac{8\pi}{13}=-\cos\frac{5\pi}{13}.
\displaystyle \therefore 2\cos\frac{\pi}{13}\cos\frac{9\pi}{13}+\cos\frac{3\pi}{13}+\cos\frac{5\pi}{13}=0.
\displaystyle \therefore \text{The required value is }0.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Evaluate the value of }\sin105^\circ
\displaystyle \text{Answer:}
\displaystyle \sin105^\circ=\sin(60^\circ+45^\circ).
\displaystyle =\sin60^\circ\cos45^\circ+\cos60^\circ\sin45^\circ.
\displaystyle =\frac{\sqrt{3}}{2}\cdot\frac{1}{\sqrt{2}}+\frac{1}{2}\cdot\frac{1}{\sqrt{2}}.
\displaystyle =\frac{\sqrt{6}+\sqrt{2}}{4}.
\displaystyle \therefore \text{The required value is }\frac{\sqrt{6}+\sqrt{2}}{4}.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Find the value of }k\text{ if }\tan A-\cot A=k\cot2A
\displaystyle \text{Answer:}
\displaystyle \tan A-\cot A=\frac{\sin^2A-\cos^2A}{\sin A\cos A}.
\displaystyle =\frac{-\cos2A}{\frac{1}{2}\sin2A}=-2\cot2A.
\displaystyle \therefore k=-2.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{Convert }1\text{ radian in degree, minute and seconds}
\displaystyle \text{Answer:}
\displaystyle 1\text{ radian}=\frac{180^\circ}{\pi}.
\displaystyle \text{Taking }\pi=\frac{22}{7},\quad 1\text{ radian}=\frac{180\times7}{22}^\circ=57.2727^\circ.
\displaystyle 0.2727^\circ\times60=16.362'\quad\text{and}\quad0.362'\times60\approx21.8''.
\displaystyle \therefore 1\text{ radian}\approx57^\circ16'22''.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Convert }40^\circ20'\text{ into radian measure}
\displaystyle \text{Answer:}
\displaystyle 40^\circ20'=40^\circ+\frac{20}{60}^\circ=\frac{121}{3}^\circ.
\displaystyle \text{Since }1^\circ=\frac{\pi}{180}\text{ radian,}
\displaystyle \frac{121}{3}^\circ=\frac{121}{3}\times\frac{\pi}{180}=\frac{121\pi}{540}\text{ radians}.
\displaystyle \therefore \text{The required radian measure is }\frac{121\pi}{540}.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{In a circle of diameter }56\text{ cm, the length of the chord is }28\text{ cm, find the}
\displaystyle \text{length of the minor arc.}
\displaystyle \text{Answer:}
\displaystyle \text{Radius of the circle }r=\frac{56}{2}=28\text{ cm}.
\displaystyle \text{Let }\theta\text{ be the angle subtended by the chord at the centre.}
\displaystyle 28=2(28)\sin\frac{\theta}{2}.
\displaystyle \sin\frac{\theta}{2}=\frac{1}{2}\implies\frac{\theta}{2}=30^\circ\implies\theta=60^\circ=\frac{\pi}{3}.
\displaystyle \text{Length of minor arc }=r\theta=28\times\frac{\pi}{3}=\frac{28\pi}{3}\text{ cm}.
\displaystyle \therefore \text{The length of the minor arc is }\frac{28\pi}{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{A wheel makes }540\text{ revolutions in one minute. Through how many radians}
\displaystyle \text{does it turn in }12\text{ seconds.}
\displaystyle \text{Answer:}
\displaystyle \text{Number of revolutions in }12\text{ seconds}=\frac{540}{60}\times12=108.
\displaystyle \text{One revolution}=2\pi\text{ radians}.
\displaystyle \therefore \text{Angle turned}=108\times2\pi=216\pi\text{ radians}.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{Find the value of }2\sin75^\circ\sin15^\circ.
\displaystyle \text{Answer:}
\displaystyle \text{Using }2\sin A\sin B=\cos(A-B)-\cos(A+B),
\displaystyle 2\sin75^\circ\sin15^\circ=\cos60^\circ-\cos90^\circ.
\displaystyle =\frac{1}{2}-0=\frac{1}{2}.
\displaystyle \therefore \text{The required value is }\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{Find }a\text{ if }\sin(-420^\circ)\cos390^\circ+\cos(-660^\circ)\sin330^\circ=-a
\displaystyle \text{Answer:}
\displaystyle \sin(-420^\circ)=\sin(-60^\circ)=-\frac{\sqrt{3}}{2},\quad\cos390^\circ=\cos30^\circ=\frac{\sqrt{3}}{2}.
\displaystyle \cos(-660^\circ)=\cos60^\circ=\frac{1}{2},\quad\sin330^\circ=-\frac{1}{2}.
\displaystyle \therefore \sin(-420^\circ)\cos390^\circ+\cos(-660^\circ)\sin330^\circ
\displaystyle =-\frac{3}{4}-\frac{1}{4}=-1.
\displaystyle \therefore -a=-1\implies a=1.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{If }\sin x+\mathrm{cosec}\,x=2,\text{ then find }\sin^8x+\mathrm{cosec}^8x
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sin x=t.\text{ Then }t+\frac{1}{t}=2.
\displaystyle t^2-2t+1=0\implies(t-1)^2=0\implies t=1.
\displaystyle \therefore \sin x=1\text{ and }\mathrm{cosec}\,x=1.
\displaystyle \therefore \sin^8x+\mathrm{cosec}^8x=1+1=2.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{What will be the domain and range of }f(x)=\sin\frac{x}{2}
\displaystyle \text{Answer:}
\displaystyle \sin\frac{x}{2}\text{ is defined for every real value of }x.
\displaystyle \therefore \text{Domain}=R.
\displaystyle \text{Also, }-1\leq\sin\frac{x}{2}\leq1.
\displaystyle \therefore \text{Range}=[-1,1].
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{Evaluate the value of }\sin(-1140^\circ)+\cos(-450^\circ)
\displaystyle \text{Answer:}
\displaystyle \sin(-1140^\circ)=\sin(-60^\circ)=-\frac{\sqrt{3}}{2}.
\displaystyle \cos(-450^\circ)=\cos(-90^\circ)=0.
\displaystyle \therefore \sin(-1140^\circ)+\cos(-450^\circ)=-\frac{\sqrt{3}}{2}+0.
\displaystyle \therefore \text{The required value is }-\frac{\sqrt{3}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find the value of }\tan15^\circ\cdot\tan45^\circ\cdot\tan75^\circ
\displaystyle \text{Answer:}
\displaystyle \tan15^\circ\tan75^\circ=\tan15^\circ\cot15^\circ=1.
\displaystyle \text{Also, }\tan45^\circ=1.
\displaystyle \therefore \tan15^\circ\cdot\tan45^\circ\cdot\tan75^\circ=1.
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \text{SHORT QUESTIONS}

\displaystyle \textbf{Question 51: }\text{If }x=y\cos\frac{2\pi}{3}=z\cos\frac{4\pi}{3},\text{ then find the value of }xy+yz+zx
\displaystyle \text{Answer:}
\displaystyle \cos\frac{2\pi}{3}=-\frac{1}{2}\quad\text{and}\quad\cos\frac{4\pi}{3}=-\frac{1}{2}.
\displaystyle \therefore x=-\frac{y}{2}=-\frac{z}{2}.
\displaystyle \therefore y=-2x\quad\text{and}\quad z=-2x.
\displaystyle xy+yz+zx=x(-2x)+(-2x)(-2x)+(-2x)x.
\displaystyle =-2x^2+4x^2-2x^2=0.
\displaystyle \therefore xy+yz+zx=0.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{Prove that }\sin(45^\circ+A)\sin(45^\circ-A)=\frac{\cos2A}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Using }2\sin P\sin Q=\cos(P-Q)-\cos(P+Q),
\displaystyle 2\sin(45^\circ+A)\sin(45^\circ-A)
\displaystyle =\cos\{(45^\circ+A)-(45^\circ-A)\}-\cos\{(45^\circ+A)+(45^\circ-A)\}.
\displaystyle =\cos2A-\cos90^\circ=\cos2A.
\displaystyle \therefore \sin(45^\circ+A)\sin(45^\circ-A)=\frac{\cos2A}{2}.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{Prove that : }\cot x\cot2x-\cot2x\cot3x-\cot3x\cot x=1
\displaystyle \text{Answer:}
\displaystyle \text{Since }3x=x+2x,\text{ we use }\cot(A+B)=\frac{\cot A\cot B-1}{\cot A+\cot B}.
\displaystyle \therefore \cot3x=\frac{\cot x\cot2x-1}{\cot x+\cot2x}.
\displaystyle \cot3x(\cot x+\cot2x)=\cot x\cot2x-1.
\displaystyle \therefore \cot x\cot2x-\cot2x\cot3x-\cot3x\cot x=1.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{If }A+B=45^\circ,\text{ then }(\cot A-1)(\cot B-1)\text{ is equal to...............}
\displaystyle \text{Answer:}
\displaystyle \tan(A+B)=\tan45^\circ=1.
\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}=1.
\displaystyle 1-\tan A-\tan B+\tan A\tan B=0.
\displaystyle (1-\tan A)(1-\tan B)=0.
\displaystyle \therefore (\cot A-1)(\cot B-1)=0.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Find }\sin\frac{x}{2},\cos\frac{x}{2}\text{ and }\tan\frac{x}{2}\text{ if }\tan x=-\frac{3}{4},
\displaystyle x\text{ lies in the 2nd quadrant}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\sin x=\frac{3}{5}\text{ and }\cos x=-\frac{4}{5}.
\displaystyle \text{Also, }\frac{x}{2}\text{ lies in the first quadrant, so all half-angle ratios are positive.}
\displaystyle \sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{1+\frac{4}{5}}{2}}=\frac{3}{\sqrt{10}}.
\displaystyle \cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\sqrt{\frac{1-\frac{4}{5}}{2}}=\frac{1}{\sqrt{10}}.
\displaystyle \tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=3.
\displaystyle \therefore \sin\frac{x}{2}=\frac{3}{\sqrt{10}},\quad\cos\frac{x}{2}=\frac{1}{\sqrt{10}},\quad\tan\frac{x}{2}=3.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Evaluate: }\sin\frac{\pi}{8}
\displaystyle \text{Answer:}
\displaystyle \sin\frac{\pi}{8}=\sqrt{\frac{1-\cos\frac{\pi}{4}}{2}}.
\displaystyle =\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}.
\displaystyle \therefore \text{The required value is }\frac{\sqrt{2-\sqrt{2}}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{If }A+B=45^\circ,\text{ then }(\cot A-1)(\cot B-1)\text{ is equal to...............}
\displaystyle \text{Answer:}
\displaystyle \cot(A+B)=\cot45^\circ=1.
\displaystyle \frac{\cot A\cot B-1}{\cot A+\cot B}=1.
\displaystyle \therefore \cot A\cot B-\cot A-\cot B=1.
\displaystyle (\cot A-1)(\cot B-1)
\displaystyle =\cot A\cot B-\cot A-\cot B+1=1+1=2.
\displaystyle \therefore \text{The required value is }2.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Find }\sin\frac{x}{2},\cos\frac{x}{2}\text{ and }\tan\frac{x}{2}\text{ if }\tan x=-\frac{3}{4},
\displaystyle x\text{ lies in the 2nd quadrant}
\displaystyle \text{Answer:}
\displaystyle \text{Since }x\text{ lies in the second quadrant, }\sin x=\frac{3}{5}\text{ and }\cos x=-\frac{4}{5}.
\displaystyle \text{Also, }\frac{x}{2}\text{ lies in the first quadrant, so all the ratios are positive.}
\displaystyle \sin\frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}=\sqrt{\frac{1+\frac{4}{5}}{2}}=\frac{3}{\sqrt{10}}.
\displaystyle \cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}=\sqrt{\frac{1-\frac{4}{5}}{2}}=\frac{1}{\sqrt{10}}.
\displaystyle \tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=3.
\displaystyle \therefore \sin\frac{x}{2}=\frac{3}{\sqrt{10}},\quad\cos\frac{x}{2}=\frac{1}{\sqrt{10}},\quad\tan\frac{x}{2}=3.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Evaluate: }\sin\frac{\pi}{8}
\displaystyle \text{Answer:}
\displaystyle \sin\frac{\pi}{8}=\sin\frac{\pi/4}{2}.
\displaystyle =\sqrt{\frac{1-\cos\frac{\pi}{4}}{2}}.
\displaystyle =\sqrt{\frac{1-\frac{1}{\sqrt{2}}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}.
\displaystyle \therefore \text{The required value is }\frac{\sqrt{2-\sqrt{2}}}{2}.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{If }\tan\left(\frac{\pi}{4}+\theta\right)-\tan\left(\frac{\pi}{4}-\theta\right)=k\tan2\theta,\text{ find }k
\displaystyle \text{Answer:}
\displaystyle \tan\left(\frac{\pi}{4}+\theta\right)=\frac{1+\tan\theta}{1-\tan\theta}.
\displaystyle \tan\left(\frac{\pi}{4}-\theta\right)=\frac{1-\tan\theta}{1+\tan\theta}.
\displaystyle \therefore \tan\left(\frac{\pi}{4}+\theta\right)-\tan\left(\frac{\pi}{4}-\theta\right)
\displaystyle =\frac{(1+\tan\theta)^2-(1-\tan\theta)^2}{1-\tan^2\theta}.
\displaystyle =\frac{4\tan\theta}{1-\tan^2\theta}=2\tan2\theta.
\displaystyle \therefore k=2.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Evaluate: }\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}
\displaystyle \text{Answer:}
\displaystyle \text{Using the identity }4\cos x\cos2x\cos4x=\frac{\sin8x}{2\sin x},
\displaystyle 4\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}
\displaystyle =\frac{\sin\frac{8\pi}{7}}{2\sin\frac{\pi}{7}}.
\displaystyle =\frac{-\sin\frac{\pi}{7}}{2\sin\frac{\pi}{7}}=-\frac{1}{2}.
\displaystyle \therefore \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}=-\frac{1}{8}.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Prove that: }\frac{\sin A-\sin B}{\cos B-\cos A}=\cot\frac{A+B}{2}
\displaystyle \text{Answer:}
\displaystyle \text{Using }\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \text{and }\cos B-\cos A=2\sin\frac{A+B}{2}\sin\frac{A-B}{2},
\displaystyle \frac{\sin A-\sin B}{\cos B-\cos A}
\displaystyle =\frac{2\cos\frac{A+B}{2}\sin\frac{A-B}{2}}{2\sin\frac{A+B}{2}\sin\frac{A-B}{2}}.
\displaystyle =\frac{\cos\frac{A+B}{2}}{\sin\frac{A+B}{2}}=\cot\frac{A+B}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Evaluate: }\sin12^\circ\cdot\sin48^\circ\cdot\sin54^\circ
\displaystyle \text{Answer:}
\displaystyle \sin54^\circ=\cos36^\circ.
\displaystyle \text{Using the identity }\sin18^\circ\sin54^\circ=\frac{1}{4},
\displaystyle \text{and }\sin12^\circ\sin48^\circ=\frac{\cos36^\circ-\cos60^\circ}{2}.
\displaystyle \sin12^\circ\sin48^\circ=\frac{\cos36^\circ-\frac{1}{2}}{2}.
\displaystyle \text{Using }\cos36^\circ=\frac{\sqrt{5}+1}{4},\text{ the product simplifies to }\frac{1}{8}.
\displaystyle \therefore \sin12^\circ\cdot\sin48^\circ\cdot\sin54^\circ=\frac{1}{8}.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{Evaluate the value of }\tan2A\text{ if }\sin A=-\frac{3}{5}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\sin A=-\frac{3}{5}.
\displaystyle \cos A=\pm\sqrt{1-\sin^2A}=\pm\frac{4}{5}.
\displaystyle \therefore \tan A=\frac{\sin A}{\cos A}=\mp\frac{3}{4}.
\displaystyle \tan2A=\frac{2\tan A}{1-\tan^2A}.
\displaystyle =\frac{2\left(\mp\frac{3}{4}\right)}{1-\frac{9}{16}}=\mp\frac{24}{7}.
\displaystyle \therefore \tan2A=\pm\frac{24}{7},\text{ depending on the quadrant of }A.
\displaystyle \text{Hence, a unique value cannot be determined without the quadrant of }A.
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Evaluate the value of }\sin75^\circ-\cos75^\circ
\displaystyle \text{Answer:}
\displaystyle \sin75^\circ-\cos75^\circ
\displaystyle =\sqrt{2}\sin(75^\circ-45^\circ).
\displaystyle =\sqrt{2}\sin30^\circ=\frac{\sqrt{2}}{2}.
\displaystyle \therefore \text{The required value is }\frac{1}{\sqrt{2}}.
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{Prove that: }\frac{\sin2A}{1+\cos2A}=\tan A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin2A}{1+\cos2A}.
\displaystyle =\frac{2\sin A\cos A}{1+\cos^2A-\sin^2A}.
\displaystyle =\frac{2\sin A\cos A}{2\cos^2A}.
\displaystyle =\frac{\sin A}{\cos A}=\tan A=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{If }\frac{\sin A+\sin2A}{1+\cos A+\cos2A}=k\tan A,\text{ find }k
\displaystyle \text{Answer:}
\displaystyle \frac{\sin A+\sin2A}{1+\cos A+\cos2A}
\displaystyle =\frac{\sin A+2\sin A\cos A}{1+\cos A+(2\cos^2A-1)}.
\displaystyle =\frac{\sin A(1+2\cos A)}{\cos A(1+2\cos A)}.
\displaystyle =\frac{\sin A}{\cos A}=\tan A.
\displaystyle \therefore k\tan A=\tan A.
\displaystyle \therefore k=1.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Evaluate the value of }(\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x
\displaystyle \text{Answer:}
\displaystyle (\sin3x+\sin x)\sin x+(\cos3x-\cos x)\cos x
\displaystyle =(2\sin2x\cos x)\sin x+(-2\sin2x\sin x)\cos x.
\displaystyle =2\sin2x\sin x\cos x-2\sin2x\sin x\cos x=0.
\displaystyle \therefore \text{The required value is }0.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{The railway train is travelling on a circular curve of }1500\text{ metres}
\displaystyle \text{radius at the rate of }66\text{ km/hr. Through what angle it has turned in }10\text{ seconds.}
\displaystyle \text{Answer:}
\displaystyle \text{Speed}=66\times\frac{5}{18}=\frac{55}{3}\text{ m/s}.
\displaystyle \text{Distance travelled in }10\text{ seconds}=\frac{55}{3}\times10=\frac{550}{3}\text{ m}.
\displaystyle \text{Using }l=r\theta,\text{ we get }\theta=\frac{l}{r}.
\displaystyle \theta=\frac{\frac{550}{3}}{1500}=\frac{11}{90}\text{ radian}.
\displaystyle \therefore \text{The train has turned through an angle of }\frac{11}{90}\text{ radian}.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{Evaluate: }\frac{\sin(180^\circ+\theta)\cos(90^\circ+\theta)\tan(270^\circ-\theta)\cot(360^\circ-\theta)}{\sin(360^\circ-\theta)\cos(360^\circ+\theta)\mathrm{cosec}(-\theta)\sin(270^\circ-\theta)}
\displaystyle \text{Answer:}
\displaystyle \sin(180^\circ+\theta)=-\sin\theta,\quad\cos(90^\circ+\theta)=-\sin\theta.
\displaystyle \tan(270^\circ-\theta)=\cot\theta,\quad\cot(360^\circ-\theta)=-\cot\theta.
\displaystyle \sin(360^\circ-\theta)=-\sin\theta,\quad\cos(360^\circ+\theta)=\cos\theta.
\displaystyle \mathrm{cosec}(-\theta)=-\mathrm{cosec}\,\theta,\quad\sin(270^\circ-\theta)=-\cos\theta.
\displaystyle \therefore \frac{(-\sin\theta)(-\sin\theta)(\cot\theta)(-\cot\theta)}{(-\sin\theta)(\cos\theta)(-\mathrm{cosec}\,\theta)(-\cos\theta)}
\displaystyle =\frac{-\cos^2\theta}{-\cos^2\theta}=1.
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{If }\sin A=-3/5\text{ and }A\text{ lies in 3rd Quadrant and }\cos B=12/13
\displaystyle \text{and }B\text{ lies in 4th Quadrant, then find the value of }\tan(A+B)
\displaystyle \text{Answer:}
\displaystyle \text{Since }A\text{ lies in the third quadrant, }\cos A=-\frac{4}{5}.
\displaystyle \therefore \tan A=\frac{\sin A}{\cos A}=\frac{-3/5}{-4/5}=\frac{3}{4}.
\displaystyle \text{Since }B\text{ lies in the fourth quadrant, }\sin B=-\frac{5}{13}.
\displaystyle \therefore \tan B=\frac{\sin B}{\cos B}=\frac{-5/13}{12/13}=-\frac{5}{12}.
\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.
\displaystyle =\frac{\frac{3}{4}-\frac{5}{12}}{1-\left(\frac{3}{4}\right)\left(-\frac{5}{12}\right)}.
\displaystyle =\frac{\frac{1}{3}}{\frac{21}{16}}=\frac{16}{63}.
\displaystyle \therefore \tan(A+B)=\frac{16}{63}.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{In any cyclic quadrilateral }ABCD,\text{ prove that}
\displaystyle \cos A+\cos B+\cos C+\cos D=0
\displaystyle \text{Answer:}
\displaystyle \text{In a cyclic quadrilateral, opposite angles are supplementary.}
\displaystyle \therefore A+C=180^\circ\quad\text{and}\quad B+D=180^\circ.
\displaystyle C=180^\circ-A\quad\text{and}\quad D=180^\circ-B.
\displaystyle \therefore \cos C=\cos(180^\circ-A)=-\cos A.
\displaystyle \text{Similarly, }\cos D=\cos(180^\circ-B)=-\cos B.
\displaystyle \therefore \cos A+\cos B+\cos C+\cos D
\displaystyle =\cos A+\cos B-\cos A-\cos B=0.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{Show that }\tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x.
\displaystyle \text{Answer:}
\displaystyle \text{Since }3x=2x+x,\text{ we use the tangent addition formula.}
\displaystyle \tan3x=\tan(2x+x)=\frac{\tan2x+\tan x}{1-\tan2x\tan x}.
\displaystyle \tan3x(1-\tan2x\tan x)=\tan2x+\tan x.
\displaystyle \tan3x-\tan3x\tan2x\tan x=\tan2x+\tan x.
\displaystyle \therefore \tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{LONG QUESTIONS}

\displaystyle \textbf{Question 71: }\text{Prove that : }\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)
\displaystyle \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)=\frac{1}{8}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right)
\displaystyle \qquad\left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right).
\displaystyle \text{We know that }\cos(\pi-\theta)=-\cos\theta.
\displaystyle \therefore \cos\frac{5\pi}{8}=-\cos\frac{3\pi}{8}\quad\text{and}\quad
\displaystyle \cos\frac{7\pi}{8}=-\cos\frac{\pi}{8}.
\displaystyle \therefore \text{LHS}=\left(1+\cos\frac{\pi}{8}\right)\left(1-\cos\frac{\pi}{8}\right)
\displaystyle \qquad\times\left(1+\cos\frac{3\pi}{8}\right)\left(1-\cos\frac{3\pi}{8}\right).
\displaystyle =\left(1-\cos^2\frac{\pi}{8}\right)\left(1-\cos^2\frac{3\pi}{8}\right).
\displaystyle =\sin^2\frac{\pi}{8}\sin^2\frac{3\pi}{8}.
\displaystyle \text{Since }\sin\frac{3\pi}{8}=\cos\frac{\pi}{8},\text{ we get}
\displaystyle \text{LHS}=\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}.
\displaystyle =\left(\sin\frac{\pi}{8}\cos\frac{\pi}{8}\right)^2.
\displaystyle =\left(\frac{1}{2}\sin\frac{\pi}{4}\right)^2.
\displaystyle =\left(\frac{1}{2}\cdot\frac{1}{\sqrt{2}}\right)^2=\frac{1}{8}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{Prove that : }\sin^4\frac{\pi}{8}+\sin^4\frac{3\pi}{8}+\sin^4\frac{5\pi}{8}
\displaystyle +\sin^4\frac{7\pi}{8}=\frac{3}{2}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin^4\frac{\pi}{8}+\sin^4\frac{3\pi}{8}+\sin^4\frac{5\pi}{8}
\displaystyle \qquad+\sin^4\frac{7\pi}{8}.
\displaystyle \text{We know that }\sin(\pi-\theta)=\sin\theta.
\displaystyle \therefore \sin\frac{5\pi}{8}=\sin\frac{3\pi}{8}\quad\text{and}\quad
\displaystyle \sin\frac{7\pi}{8}=\sin\frac{\pi}{8}.
\displaystyle \therefore \text{LHS}=2\left(\sin^4\frac{\pi}{8}+\sin^4\frac{3\pi}{8}\right).
\displaystyle \text{Also, }\sin\frac{3\pi}{8}=\cos\frac{\pi}{8}.
\displaystyle \therefore \text{LHS}=2\left(\sin^4\frac{\pi}{8}+\cos^4\frac{\pi}{8}\right).
\displaystyle \text{Using }\sin^4\theta+\cos^4\theta=(\sin^2\theta+\cos^2\theta)^2
\displaystyle \qquad-2\sin^2\theta\cos^2\theta,
\displaystyle \text{LHS}=2\left(1-2\sin^2\frac{\pi}{8}\cos^2\frac{\pi}{8}\right).
\displaystyle =2\left(1-\frac{1}{2}\sin^2\frac{\pi}{4}\right).
\displaystyle =2\left(1-\frac{1}{2}\cdot\frac{1}{2}\right).
\displaystyle =2\left(1-\frac{1}{4}\right)=2\cdot\frac{3}{4}=\frac{3}{2}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{Prove that : }\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ.
\displaystyle \text{Using }\sin3A=4\sin A\sin(60^\circ+A)\sin(60^\circ-A),
\displaystyle \text{with }A=20^\circ,\text{ we get}
\displaystyle \sin60^\circ=4\sin20^\circ\sin80^\circ\sin40^\circ.
\displaystyle \therefore \sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sin60^\circ}{4}.
\displaystyle \therefore \text{LHS}=\frac{\sin60^\circ}{4}\cdot\sin60^\circ.
\displaystyle =\frac{1}{4}\sin^260^\circ.
\displaystyle =\frac{1}{4}\left(\frac{\sqrt{3}}{2}\right)^2=\frac{3}{16}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 74(a): }\text{Prove that: }\cos^2\alpha+\cos^2(\alpha+120^\circ)+\cos^2(\alpha-120^\circ)=\frac{3}{2}
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\cos^2\alpha+\cos^2(\alpha+120^\circ)+\cos^2(\alpha-120^\circ).
\displaystyle \text{Using }\cos^2\theta=\frac{1+\cos2\theta}{2},
\displaystyle \text{LHS}=\frac{1+\cos2\alpha}{2}+\frac{1+\cos(2\alpha+240^\circ)}{2}
\displaystyle \qquad+\frac{1+\cos(2\alpha-240^\circ)}{2}.
\displaystyle =\frac{3}{2}+\frac{1}{2}\{\cos2\alpha+\cos(2\alpha+240^\circ)
\displaystyle \qquad+\cos(2\alpha-240^\circ)\}.
\displaystyle \cos(2\alpha+240^\circ)+\cos(2\alpha-240^\circ)
\displaystyle =2\cos2\alpha\cos240^\circ=-\cos2\alpha.
\displaystyle \therefore \cos2\alpha+\cos(2\alpha+240^\circ)+\cos(2\alpha-240^\circ)=0.
\displaystyle \therefore \text{LHS}=\frac{3}{2}.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 74(b): }\text{Prove that : }\frac{\sin(4A-2B)+\sin(4B-2A)}{\cos(4A-2B)+\cos(4A-2B)}=\tan(A+B)
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{\sin(4A-2B)+\sin(4B-2A)}{\cos(4A-2B)+\cos(4B-2A)}.
\displaystyle \text{Using }\sin C+\sin D=2\sin\frac{C+D}{2}\cos\frac{C-D}{2},
\displaystyle \sin(4A-2B)+\sin(4B-2A)=2\sin(A+B)\cos3(A-B).
\displaystyle \text{Also, using }\cos C+\cos D=2\cos\frac{C+D}{2}\cos\frac{C-D}{2},
\displaystyle \cos(4A-2B)+\cos(4B-2A)=2\cos(A+B)\cos3(A-B).
\displaystyle \therefore \text{LHS}=\frac{2\sin(A+B)\cos3(A-B)}{2\cos(A+B)\cos3(A-B)}.
\displaystyle =\frac{\sin(A+B)}{\cos(A+B)}=\tan(A+B)=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{Prove that : }\frac{\sin A-\sin5A+\sin9A-\sin13A}{\cos A-\cos5A-\cos9A+\cos13A}=\cot4A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\frac{(\sin A-\sin5A)+(\sin9A-\sin13A)}{(\cos A-\cos5A)+(\cos13A-\cos9A)}.
\displaystyle \text{Using }\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2},
\displaystyle \sin A-\sin5A=-2\sin2A\cos3A.
\displaystyle \sin9A-\sin13A=-2\sin2A\cos11A.
\displaystyle \therefore \text{Numerator}=-2\sin2A(\cos3A+\cos11A).
\displaystyle =-4\sin2A\cos7A\cos4A.
\displaystyle \text{Using }\cos C-\cos D=-2\sin\frac{C+D}{2}\sin\frac{C-D}{2},
\displaystyle \cos A-\cos5A=2\sin3A\sin2A.
\displaystyle \cos13A-\cos9A=-2\sin11A\sin2A.
\displaystyle \therefore \text{Denominator}=2\sin2A(\sin3A-\sin11A).
\displaystyle =-4\sin2A\cos7A\sin4A.
\displaystyle \therefore \text{LHS}=\frac{-4\sin2A\cos7A\cos4A}{-4\sin2A\cos7A\sin4A}.
\displaystyle =\frac{\cos4A}{\sin4A}=\cot4A=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{Prove that : }\sqrt{2+\sqrt{2+\sqrt{2+2\cos8A}}}=2\cos A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sqrt{2+\sqrt{2+\sqrt{2+2\cos8A}}}.
\displaystyle \text{Using }1+\cos2\theta=2\cos^2\theta,\text{ we get}
\displaystyle \sqrt{2+2\cos8A}=\sqrt{4\cos^24A}=2\cos4A.
\displaystyle \therefore \text{LHS}=\sqrt{2+\sqrt{2+2\cos4A}}.
\displaystyle \text{Again, }2+2\cos4A=4\cos^22A.
\displaystyle \therefore \sqrt{2+2\cos4A}=2\cos2A.
\displaystyle \therefore \text{LHS}=\sqrt{2+2\cos2A}.
\displaystyle =\sqrt{4\cos^2A}=2\cos A.
\displaystyle \therefore \text{LHS}=\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{Prove that : }\sin4A=4\sin A\cos^3A-4\cos A\sin^3A
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=\sin4A.
\displaystyle =2\sin2A\cos2A.
\displaystyle =2(2\sin A\cos A)(\cos^2A-\sin^2A).
\displaystyle =4\sin A\cos A(\cos^2A-\sin^2A).
\displaystyle =4\sin A\cos^3A-4\cos A\sin^3A.
\displaystyle =\text{RHS}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 78: }\text{If }A+B+C=\pi,\text{ then prove that:}
\displaystyle \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1
\displaystyle \text{Answer:}
\displaystyle \text{Given, }A+B+C=\pi.
\displaystyle \therefore \frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2}.
\displaystyle \text{Using the formula for }\tan(x+y+z),
\displaystyle \tan(x+y+z)=\frac{\tan x+\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan y\tan z-\tan z\tan x}.
\displaystyle \text{Since }\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi}{2},\text{ its tangent is not defined.}
\displaystyle \therefore 1-\tan\frac{A}{2}\tan\frac{B}{2}-\tan\frac{B}{2}\tan\frac{C}{2}
\displaystyle \qquad-\tan\frac{C}{2}\tan\frac{A}{2}=0.
\displaystyle \therefore \tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{Prove that }2\sin^2\beta+4\cos(\alpha+\beta)\sin\alpha\sin\beta+\cos2(\alpha+\beta)
\displaystyle \text{is independent of }\beta.
\displaystyle \text{Answer:}
\displaystyle \text{Let }E=2\sin^2\beta+4\cos(\alpha+\beta)\sin\alpha\sin\beta+\cos2(\alpha+\beta).
\displaystyle \text{Using }\cos2(\alpha+\beta)=1-2\sin^2(\alpha+\beta),\text{ we get}
\displaystyle E=2\sin^2\beta+4\cos(\alpha+\beta)\sin\alpha\sin\beta+1-2\sin^2(\alpha+\beta).
\displaystyle \text{Now, }\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta.
\displaystyle \text{So, }\sin^2(\alpha+\beta)=\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta
\displaystyle \qquad+2\sin\alpha\cos\alpha\sin\beta\cos\beta.
\displaystyle \text{Hence,}
\displaystyle E=2\sin^2\beta+4\sin\alpha\sin\beta\cos(\alpha+\beta)+1
\displaystyle \qquad-2[\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta+2\sin\alpha\cos\alpha\sin\beta\cos\beta].
\displaystyle =2\sin^2\beta+4\sin\alpha\sin\beta(\cos\alpha\cos\beta-\sin\alpha\sin\beta)+1
\displaystyle \qquad-2\sin^2\alpha\cos^2\beta-2\cos^2\alpha\sin^2\beta-4\sin\alpha\cos\alpha\sin\beta\cos\beta.
\displaystyle =2\sin^2\beta+4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+1
\displaystyle \qquad-2\sin^2\alpha\cos^2\beta-2\cos^2\alpha\sin^2\beta-4\sin\alpha\cos\alpha\sin\beta\cos\beta.
\displaystyle =2\sin^2\beta-4\sin^2\alpha\sin^2\beta+1-2\sin^2\alpha\cos^2\beta-2\cos^2\alpha\sin^2\beta.
\displaystyle =1-2\sin^2\alpha(\cos^2\beta+\sin^2\beta).
\displaystyle =1-2\sin^2\alpha.
\displaystyle =\cos2\alpha.
\displaystyle \text{Thus, the expression is independent of }\beta\text{ (its value is }\cos2\alpha\text{).}
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Sketch the graph of }\cos2x\text{ and }\cos\left(2x-\frac{\pi}{4}\right).
\displaystyle \text{Answer:}
\displaystyle \text{The graphs of }y=\cos2x\text{ and }y=\cos\left(2x-\frac{\pi}{4}\right)\text{ are cosine curves}
\displaystyle \text{of the same amplitude and period.}
\displaystyle \text{Amplitude}=1,\qquad \text{Period}=\pi.
\displaystyle \text{The graph of }y=\cos\left(2x-\frac{\pi}{4}\right)\text{ is the graph of }y=\cos2x
\displaystyle \text{shifted to the right by }\frac{\pi}{8}.
\displaystyle \\

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