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

\displaystyle \textbf{Question 1: }\text{If }\tan\theta=x-\frac{1}{4x},\text{ then }\sec\theta-\tan\theta\text{ is equal to}
\displaystyle \text{(a) }-2x,\ \frac{1}{2x}\qquad\text{(b) }-\frac{1}{2x},\ 2x
\displaystyle \text{(c) }2x\qquad\text{(d) }2x,\ \frac{1}{2x}
\displaystyle \text{Answer:}
\displaystyle \sec^2\theta-\tan^2\theta=1
\displaystyle \Rightarrow \sec^2\theta=1+\left(x-\frac{1}{4x}\right)^2
\displaystyle =x^2+\frac12+\frac{1}{16x^2}=\left(x+\frac{1}{4x}\right)^2.
\displaystyle \therefore \sec\theta=\pm\left(x+\frac{1}{4x}\right).
\displaystyle \text{If }\sec\theta=x+\frac{1}{4x},
\displaystyle \sec\theta-\tan\theta=\frac{1}{2x}.
\displaystyle \text{If }\sec\theta=-x-\frac{1}{4x},
\displaystyle \sec\theta-\tan\theta=-2x.
\displaystyle \therefore \sec\theta-\tan\theta=-2x\text{ or }\frac{1}{2x}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\sec\theta=x+\frac{1}{4x},\text{ then }\sec\theta+\tan\theta=
\displaystyle \text{(a) }x,\ \frac1x\qquad\text{(b) }2x,\ \frac{1}{2x}
\displaystyle \text{(c) }-2x,\ \frac{1}{2x}\qquad\text{(d) }-\frac1x,\ x
\displaystyle \text{Answer:}
\displaystyle \sec^2\theta-\tan^2\theta=1
\displaystyle \Rightarrow \tan^2\theta=\left(x+\frac{1}{4x}\right)^2-1
\displaystyle =x^2-\frac12+\frac{1}{16x^2}=\left(x-\frac{1}{4x}\right)^2.
\displaystyle \therefore \tan\theta=\pm\left(x-\frac{1}{4x}\right).
\displaystyle \text{If }\tan\theta=x-\frac{1}{4x},
\displaystyle \sec\theta+\tan\theta=2x.
\displaystyle \text{If }\tan\theta=-x+\frac{1}{4x},
\displaystyle \sec\theta+\tan\theta=\frac{1}{2x}.
\displaystyle \therefore \sec\theta+\tan\theta=2x\text{ or }\frac{1}{2x}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\frac{\pi}{2}<\theta<\frac{3\pi}{2},\text{ then }  \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}\text{ is equal to}
\displaystyle \text{(a) }\sec\theta-\tan\theta\qquad\text{(b) }\sec\theta+\tan\theta
\displaystyle \text{(c) }\tan\theta-\sec\theta\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}
\displaystyle =\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}
\displaystyle =\sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}
\displaystyle =\frac{|1-\sin\theta|}{|\cos\theta|}.
\displaystyle \text{Since }1-\sin\theta\geq0,\quad |1-\sin\theta|=1-\sin\theta.
\displaystyle \text{Also, }\frac{\pi}{2}<\theta<\frac{3\pi}{2}\Rightarrow\cos\theta<0.
\displaystyle \therefore |\cos\theta|=-\cos\theta.
\displaystyle \therefore \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}
\displaystyle =-\frac{1-\sin\theta}{\cos\theta}
\displaystyle =\frac{\sin\theta-1}{\cos\theta}=\tan\theta-\sec\theta.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\pi<\theta<2\pi,\text{ then }\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\text{ is equal to}
\displaystyle \text{(a) }\mathrm{cosec}\,\theta+\cot\theta\qquad\text{(b) }\mathrm{cosec}\,\theta-\cot\theta
\displaystyle \text{(c) }-\mathrm{cosec}\,\theta+\cot\theta\qquad\text{(d) }-\mathrm{cosec}\,\theta-\cot\theta
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}
\displaystyle =\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}
\displaystyle =\frac{|1+\cos\theta|}{|\sin\theta|}.
\displaystyle \text{Since }1+\cos\theta\geq0,\quad |1+\cos\theta|=1+\cos\theta.
\displaystyle \pi<\theta<2\pi\Rightarrow\sin\theta<0.
\displaystyle \therefore |\sin\theta|=-\sin\theta.
\displaystyle \therefore \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}
\displaystyle =-\frac{1+\cos\theta}{\sin\theta}
\displaystyle =-\left(\frac1{\sin\theta}+\frac{\cos\theta}{\sin\theta}\right)
\displaystyle =-\mathrm{cosec}\,\theta-\cot\theta.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }0<\theta<\frac{\pi}{2}\text{ and if }\frac{y+1}{1-y}=\sqrt{\frac{1+\sin\theta}{1-\sin\theta}},   \text{ then }y\text{ is equal to}
\displaystyle \text{(a) }\cot\frac{\theta}{2}\qquad\text{(b) }\tan\frac{\theta}{2}
\displaystyle \text{(c) }\cot\frac{\theta}{2}+\tan\frac{\theta}{2}\qquad\text{(d) }\cot\frac{\theta}{2}-\tan\frac{\theta}{2}
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}
\displaystyle =\sqrt{\frac{(1+\sin\theta)^2}{1-\sin^2\theta}}
\displaystyle =\frac{1+\sin\theta}{\cos\theta}
\displaystyle =\frac{1+\tan(\theta/2)}{1-\tan(\theta/2)}.
\displaystyle \therefore \frac{y+1}{1-y}=\frac{1+\tan(\theta/2)}{1-\tan(\theta/2)}.
\displaystyle \therefore y=\tan\frac{\theta}{2}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\frac{\pi}{2}<\theta<\pi,\text{ then }   \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}+\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}\text{ is equal to}
\displaystyle \text{(a) }2\sec\theta\qquad\text{(b) }-2\sec\theta
\displaystyle \text{(c) }\sec\theta\qquad\text{(d) }-\sec\theta
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\frac{1-\sin\theta}{|\cos\theta|}
\displaystyle \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=\frac{1+\sin\theta}{|\cos\theta|}.
\displaystyle \therefore \text{Required expression}=\frac{2}{|\cos\theta|}.
\displaystyle \frac{\pi}{2}<\theta<\pi\Rightarrow\cos\theta<0.
\displaystyle \therefore |\cos\theta|=-\cos\theta.
\displaystyle \therefore \text{Required expression}=-\frac{2}{\cos\theta}=-2\sec\theta.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }x=r\sin\theta\cos\phi,\ y=r\sin\theta\sin\phi\text{ and }z=r\cos\theta,
\displaystyle \text{then }x^2+y^2+z^2\text{ is independent of}
\displaystyle \text{(a) }\theta,\phi\qquad\text{(b) }r,\theta\qquad\text{(c) }r,\phi\qquad\text{(d) }r
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+z^2
\displaystyle =r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi+r^2\cos^2\theta
\displaystyle =r^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+r^2\cos^2\theta
\displaystyle =r^2(\sin^2\theta+\cos^2\theta)=r^2.
\displaystyle \therefore x^2+y^2+z^2\text{ is independent of }\theta\text{ and }\phi.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\tan\theta+\sec\theta=\sqrt3,\ 0<\theta<\pi,\text{ then }\theta\text{ is equal to}
\displaystyle \text{(a) }\frac{5\pi}{6}\qquad\text{(b) }\frac{2\pi}{3}\qquad\text{(c) }\frac{\pi}{6}\qquad\text{(d) }\frac{\pi}{3}
\displaystyle \text{Answer:}
\displaystyle (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1.
\displaystyle \therefore \sec\theta-\tan\theta=\frac1{\sqrt3}.
\displaystyle \sec\theta+\tan\theta=\sqrt3.
\displaystyle \text{Subtracting, }2\tan\theta=\sqrt3-\frac1{\sqrt3}=\frac2{\sqrt3}.
\displaystyle \therefore \tan\theta=\frac1{\sqrt3}.
\displaystyle 0<\theta<\pi\text{ and }\sec\theta+\tan\theta>0\Rightarrow\theta=\frac{\pi}{6}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If }\tan\theta=-\frac{1}{\sqrt5}\text{ and }\theta\text{ lies in the IV quadrant,}
\displaystyle \text{then the value of }\cos\theta\text{ is}
\displaystyle \text{(a) }\frac{\sqrt5}{\sqrt6}\qquad\text{(b) }\frac2{\sqrt6}\qquad\text{(c) }\frac12\qquad\text{(d) }\frac1{\sqrt6}
\displaystyle \text{Answer:}
\displaystyle \sec^2\theta=1+\tan^2\theta
\displaystyle =1+\frac15=\frac65.
\displaystyle \therefore \sec\theta=\pm\sqrt{\frac65}.
\displaystyle \text{Since }\theta\text{ lies in the IV quadrant, }\cos\theta>0.
\displaystyle \therefore \sec\theta=\sqrt{\frac65}.
\displaystyle \therefore \cos\theta=\frac1{\sec\theta}=\sqrt{\frac56}=\frac{\sqrt5}{\sqrt6}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\frac{3\pi}{4}<\alpha<\pi,\text{ then }   \sqrt{2\cot\alpha+\frac{1}{\sin^2\alpha}}\text{ is equal to}
\displaystyle \text{(a) }1-\cot\alpha\qquad\text{(b) }1+\cot\alpha
\displaystyle \text{(c) }-1+\cot\alpha\qquad\text{(d) }-1-\cot\alpha
\displaystyle \text{Answer:}
\displaystyle \frac{1}{\sin^2\alpha}=\mathrm{cosec}^2\alpha=1+\cot^2\alpha.
\displaystyle \therefore 2\cot\alpha+\frac{1}{\sin^2\alpha}
\displaystyle =2\cot\alpha+1+\cot^2\alpha=(1+\cot\alpha)^2.
\displaystyle \therefore \sqrt{2\cot\alpha+\frac{1}{\sin^2\alpha}}=|1+\cot\alpha|.
\displaystyle \frac{3\pi}{4}<\alpha<\pi\Rightarrow \cot\alpha<-1.
\displaystyle \therefore 1+\cot\alpha<0.
\displaystyle \therefore |1+\cot\alpha|=-1-\cot\alpha.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\sin^6A+\cos^6A+3\sin^2A\cos^2A=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \sin^6A+\cos^6A
\displaystyle =(\sin^2A+\cos^2A)^3
\displaystyle \quad-3\sin^2A\cos^2A(\sin^2A+\cos^2A)
\displaystyle =1-3\sin^2A\cos^2A.
\displaystyle \therefore \sin^6A+\cos^6A+3\sin^2A\cos^2A=1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\mathrm{cosec}\,\theta-\cot\theta=\frac12,\ 0<\theta<\frac{\pi}{2},   \text{then }\cos\theta\text{ is equal to}
\displaystyle \text{(a) }\frac53\qquad\text{(b) }\frac35\qquad\text{(c) }-\frac35\qquad\text{(d) }-\frac53
\displaystyle \text{Answer:}
\displaystyle (\mathrm{cosec}\,\theta-\cot\theta)(\mathrm{cosec}\,\theta+\cot\theta)=1.
\displaystyle \therefore \frac12(\mathrm{cosec}\,\theta+\cot\theta)=1
\displaystyle \Rightarrow \mathrm{cosec}\,\theta+\cot\theta=2.
\displaystyle \mathrm{cosec}\,\theta-\cot\theta=\frac12.
\displaystyle \text{Adding, }2\mathrm{cosec}\,\theta=\frac52
\displaystyle \Rightarrow \mathrm{cosec}\,\theta=\frac54,\quad\sin\theta=\frac45.
\displaystyle \text{Subtracting, }2\cot\theta=\frac32\Rightarrow\cot\theta=\frac34.
\displaystyle \cos\theta=\sin\theta\cot\theta=\frac45\times\frac34=\frac35.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\mathrm{cosec}\,\theta+\cot\theta=\frac{11}{2},\text{ then }\tan\theta=
\displaystyle \text{(a) }\frac{21}{22}\qquad\text{(b) }\frac{15}{16}\qquad\text{(c) }\frac{44}{117}\qquad\text{(d) }\frac{117}{44}
\displaystyle \text{Answer:}
\displaystyle (\mathrm{cosec}\,\theta+\cot\theta)(\mathrm{cosec}\,\theta-\cot\theta)=1.
\displaystyle \therefore \mathrm{cosec}\,\theta-\cot\theta=\frac{2}{11}.
\displaystyle \mathrm{cosec}\,\theta+\cot\theta=\frac{11}{2}.
\displaystyle \text{Adding, }2\mathrm{cosec}\,\theta=\frac{11}{2}+\frac{2}{11}=\frac{125}{22}.
\displaystyle \therefore \mathrm{cosec}\,\theta=\frac{125}{44}.
\displaystyle \text{Subtracting, }2\cot\theta=\frac{11}{2}-\frac{2}{11}=\frac{117}{22}.
\displaystyle \therefore \cot\theta=\frac{117}{44}.
\displaystyle \therefore \tan\theta=\frac{44}{117}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\sec^2\theta=\frac{4xy}{(x+y)^2}\text{ is true if and only if}
\displaystyle \text{(a) }x+y\ne0\qquad\text{(b) }x=y,\ x\ne0
\displaystyle \text{(c) }x=y\qquad\text{(d) }x\ne0,\ y\ne0
\displaystyle \text{Answer:}
\displaystyle \sec^2\theta\geq1.
\displaystyle \therefore \frac{4xy}{(x+y)^2}\geq1.
\displaystyle \Rightarrow 4xy\geq(x+y)^2
\displaystyle \Rightarrow 4xy\geq x^2+2xy+y^2
\displaystyle \Rightarrow 0\geq x^2-2xy+y^2
\displaystyle \Rightarrow (x-y)^2\leq0.
\displaystyle \text{But }(x-y)^2\geq0,\text{ hence }x=y.
\displaystyle \text{Also, }x=y\ne0\text{ is required for the fraction to be defined.}
\displaystyle \text{For }x=y\ne0,\quad \frac{4xy}{(x+y)^2}=1=\sec^2\theta\text{ for suitable }\theta.
\displaystyle \therefore x=y,\ x\ne0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\theta\text{ is an acute angle and }\tan\theta=\frac{1}{\sqrt7},\text{ then the value of}
\displaystyle \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}\text{ is}
\displaystyle \text{(a) }\frac34\qquad\text{(b) }\frac12\qquad\text{(c) }2\qquad\text{(d) }\frac54
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac1{\sqrt7}\Rightarrow\tan^2\theta=\frac17.
\displaystyle \sec^2\theta=1+\tan^2\theta=1+\frac17=\frac87.
\displaystyle \cot^2\theta=7.
\displaystyle \therefore \mathrm{cosec}^2\theta=1+\cot^2\theta=8.
\displaystyle \therefore \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}
\displaystyle =\frac{8-\frac87}{8+\frac87}=\frac{48/7}{64/7}=\frac34.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The value of }
\displaystyle \sin^25^\circ+\sin^210^\circ+\sin^215^\circ+\cdots +\sin^285^\circ+\sin^290^\circ\text{ is} 
\displaystyle \text{(a) }7\qquad\text{(b) }8\qquad\text{(c) }9.5\qquad\text{(d) }10
\displaystyle \text{Answer:}
\displaystyle \sin^2\theta+\sin^2(90^\circ-\theta)=\sin^2\theta+\cos^2\theta=1.
\displaystyle \text{Pairing }5^\circ\text{ with }85^\circ,\ 10^\circ\text{ with }80^\circ,\ldots,40^\circ\text{ with }50^\circ,
\displaystyle \text{we obtain }8\text{ pairs, each having sum }1.
\displaystyle \text{Also, }\sin^245^\circ=\frac12\text{ and }\sin^290^\circ=1.
\displaystyle \therefore \text{Required sum}=8+\frac12+1=\frac{19}{2}=9.5.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\sin^2\frac{\pi}{18}+\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}=
\displaystyle \text{(a) }1\qquad\text{(b) }4\qquad\text{(c) }2\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \frac{\pi}{18}+\frac{4\pi}{9}=\frac{\pi}{2}
\displaystyle \text{and}\quad \frac{\pi}{9}+\frac{7\pi}{18}=\frac{\pi}{2}.
\displaystyle \therefore \sin^2\frac{\pi}{18}+\sin^2\frac{4\pi}{9}
\displaystyle =\sin^2\frac{\pi}{18}+\cos^2\frac{\pi}{18}=1.
\displaystyle \text{Also, }\sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}
\displaystyle =\sin^2\frac{\pi}{9}+\cos^2\frac{\pi}{9}=1.
\displaystyle \therefore \text{Required value}=1+1=2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\tan A+\cot A=4,\text{ then }\tan^4A+\cot^4A\text{ is equal to}
\displaystyle \text{(a) }110\qquad\text{(b) }191\qquad\text{(c) }80\qquad\text{(d) }194
\displaystyle \text{Answer:}
\displaystyle \tan A+\cot A=4.
\displaystyle \text{Squaring both sides,}
\displaystyle \tan^2A+\cot^2A+2\tan A\cot A=16.
\displaystyle \text{Since }\tan A\cot A=1,
\displaystyle \tan^2A+\cot^2A=14.
\displaystyle \text{Squaring again,}
\displaystyle \tan^4A+\cot^4A+2\tan^2A\cot^2A=196.
\displaystyle \therefore \tan^4A+\cot^4A=196-2=194.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }x\sin45^\circ\cos^260^\circ=\frac{\tan^260^\circ\mathrm{cosec}\,30^\circ}{\sec45^\circ\cot^230^\circ},
\displaystyle \text{then }x=
\displaystyle \text{(a) }2\qquad\text{(b) }4\qquad\text{(c) }8\qquad\text{(d) }16
\displaystyle \text{Answer:}
\displaystyle \sin45^\circ=\frac1{\sqrt2},\quad\cos60^\circ=\frac12,\quad\tan60^\circ=\sqrt3,
\displaystyle \mathrm{cosec}\,30^\circ=2,\quad\sec45^\circ=\sqrt2,\quad\cot30^\circ=\sqrt3.
\displaystyle \therefore x\left(\frac1{\sqrt2}\right)\left(\frac12\right)^2
\displaystyle =\frac{(\sqrt3)^2(2)}{\sqrt2(\sqrt3)^2}
\displaystyle \Rightarrow \frac{x}{4\sqrt2}=\frac{2}{\sqrt2}
\displaystyle \Rightarrow x=8.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{If }\mathrm{cosec}\,\theta-\cot\theta=\frac12,\ 0<\theta<\frac{\pi}{2},
\displaystyle \text{then }\cos\theta\text{ is equal to}
\displaystyle \text{(a) }-\frac35\qquad\text{(b) }-\frac53\qquad\text{(c) }\frac53\qquad\text{(d) }\frac35
\displaystyle \text{Answer:}
\displaystyle (\mathrm{cosec}\,\theta-\cot\theta)(\mathrm{cosec}\,\theta+\cot\theta)=1.
\displaystyle \therefore \mathrm{cosec}\,\theta+\cot\theta=2.
\displaystyle \mathrm{cosec}\,\theta-\cot\theta=\frac12.
\displaystyle \text{Adding, }2\mathrm{cosec}\,\theta=\frac52
\displaystyle \Rightarrow \mathrm{cosec}\,\theta=\frac54\Rightarrow\sin\theta=\frac45.
\displaystyle \text{Subtracting, }2\cot\theta=\frac32\Rightarrow\cot\theta=\frac34.
\displaystyle \therefore \cos\theta=\sin\theta\cot\theta=\frac45\times\frac34=\frac35.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\mathrm{cosec}\,A+\cot A=\frac{11}{2},\text{ then }\tan A=
\displaystyle \text{(a) }\frac{21}{22}\qquad\text{(b) }\frac{15}{16}\qquad\text{(c) }\frac{44}{117}\qquad\text{(d) }\frac{117}{43}
\displaystyle \text{Answer:}
\displaystyle (\mathrm{cosec}\,A+\cot A)(\mathrm{cosec}\,A-\cot A)=1.
\displaystyle \therefore \mathrm{cosec}\,A-\cot A=\frac{2}{11}.
\displaystyle \mathrm{cosec}\,A+\cot A=\frac{11}{2}.
\displaystyle \text{Subtracting, }2\cot A=\frac{11}{2}-\frac{2}{11}=\frac{117}{22}.
\displaystyle \therefore \cot A=\frac{117}{44}.
\displaystyle \therefore \tan A=\frac{44}{117}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }\tan\theta+\sec\theta=e^x,\text{ then }\cos\theta\text{ equals}
\displaystyle \text{(a) }\frac{e^x+e^{-x}}{2}\qquad\text{(b) }\frac{2}{e^x+e^{-x}}
\displaystyle \text{(c) }\frac{e^x-e^{-x}}{2}\qquad\text{(d) }\frac{e^x-e^{-x}}{e^x+e^{-x}}
\displaystyle \text{Answer:}
\displaystyle (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1.
\displaystyle \text{Since }\sec\theta+\tan\theta=e^x,
\displaystyle \therefore \sec\theta-\tan\theta=e^{-x}.
\displaystyle \text{Adding the two equations,}
\displaystyle 2\sec\theta=e^x+e^{-x}.
\displaystyle \therefore \sec\theta=\frac{e^x+e^{-x}}{2}.
\displaystyle \therefore \cos\theta=\frac{2}{e^x+e^{-x}}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\sec\theta+\tan\theta=k,\quad \cos\theta=
\displaystyle \text{(a) }\frac{k^2+1}{2k}\qquad\text{(b) }\frac{2k}{k^2+1}
\displaystyle \text{(c) }\frac{k}{k^2+1}\qquad\text{(d) }\frac{k}{k^2-1}
\displaystyle \text{Answer:}
\displaystyle (\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=1.
\displaystyle \text{Since }\sec\theta+\tan\theta=k,
\displaystyle \therefore \sec\theta-\tan\theta=\frac1k.
\displaystyle \text{Adding the two equations,}
\displaystyle 2\sec\theta=k+\frac1k=\frac{k^2+1}{k}.
\displaystyle \therefore \sec\theta=\frac{k^2+1}{2k}.
\displaystyle \therefore \cos\theta=\frac{2k}{k^2+1}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }f(x)=\cos^2x+\sec^2x,\text{ then}
\displaystyle \text{(a) }f(x)<1\qquad\text{(b) }f(x)=1
\displaystyle \text{(c) }2<f(x)<1\qquad\text{(d) }f(x)\geq2
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\cos^2x.
\displaystyle \text{Since }\sec x\text{ is defined, }0<t\leq1.
\displaystyle f(x)=t+\frac1t.
\displaystyle \text{For }t>0,\quad t+\frac1t\geq2.
\displaystyle \therefore f(x)\geq2.
\displaystyle \text{Equality occurs when }t=1,\text{ i.e., }\cos^2x=1.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Which of the following is incorrect?}
\displaystyle \text{(a) }\sin\theta=-\frac15\qquad\text{(b) }\cos\theta=1
\displaystyle \text{(c) }\sec\theta=\frac12\qquad\text{(d) }\tan\theta=20
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin\theta\leq1\text{ and }-1\leq\cos\theta\leq1.
\displaystyle \therefore \sin\theta=-\frac15\text{ and }\cos\theta=1\text{ are possible.}
\displaystyle \tan\theta\text{ can take any real value, so }\tan\theta=20\text{ is possible.}
\displaystyle \text{But }|\sec\theta|\geq1\text{ whenever }\sec\theta\text{ is defined.}
\displaystyle \therefore \sec\theta=\frac12\text{ is not possible for real }\theta.
\displaystyle \therefore \text{Option (c) is incorrect.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The value of }\cos1^\circ\cos2^\circ\cos3^\circ\cdots\cos179^\circ\text{ is}
\displaystyle \text{(a) }\frac1{\sqrt2}\qquad\text{(b) }0\qquad\text{(c) }1\qquad\text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \text{The product contains the factor }\cos90^\circ.
\displaystyle \cos90^\circ=0.
\displaystyle \therefore \cos1^\circ\cos2^\circ\cos3^\circ\cdots\cos179^\circ=0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The value of }\tan1^\circ\tan2^\circ\tan3^\circ\cdots\tan89^\circ\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }\frac12\qquad\text{(d) not defined}
\displaystyle \text{Answer:}
\displaystyle \tan(90^\circ-\theta)=\cot\theta=\frac1{\tan\theta}.
\displaystyle \therefore \tan\theta\tan(90^\circ-\theta)=1.
\displaystyle \text{Pairing }\tan1^\circ\text{ with }\tan89^\circ,\ \tan2^\circ\text{ with }\tan88^\circ,\ldots,
\displaystyle \text{each of the }44\text{ pairs has product }1.
\displaystyle \text{The remaining term is }\tan45^\circ=1.
\displaystyle \therefore \tan1^\circ\tan2^\circ\cdots\tan89^\circ=1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Which of the following is correct?}
\displaystyle \text{(a) }\sin1^\circ>\sin1\qquad\text{(b) }\sin1^\circ<\sin1
\displaystyle \text{(c) }\sin1^\circ=\sin1\qquad\text{(d) }\sin1^\circ=\frac{\pi}{180}\sin1
\displaystyle \text{Answer:}
\displaystyle 1^\circ=\frac{\pi}{180}\text{ radians}.
\displaystyle 0<\frac{\pi}{180}<1<\frac{\pi}{2}.
\displaystyle \sin x\text{ is increasing for }0<x<\frac{\pi}{2}.
\displaystyle \therefore \sin\frac{\pi}{180}<\sin1.
\displaystyle \therefore \sin1^\circ<\sin1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }A\text{ lies in second quadrant and }3\tan A+4=0,\text{ then the value of}
\displaystyle 2\cot A-5\cos A+\sin A\text{ is equal to}
\displaystyle \text{(a) }-\frac{53}{10}\qquad\text{(b) }\frac{23}{10}\qquad\text{(c) }\frac{37}{10}\qquad\text{(d) }\frac7{10}
\displaystyle \text{Answer:}
\displaystyle 3\tan A+4=0\Rightarrow\tan A=-\frac43.
\displaystyle \text{Since }A\text{ lies in the second quadrant, }\sin A>0\text{ and }\cos A<0.
\displaystyle \therefore \sin A=\frac45,\qquad\cos A=-\frac35,\qquad\cot A=-\frac34.
\displaystyle \therefore 2\cot A-5\cos A+\sin A
\displaystyle =2\left(-\frac34\right)-5\left(-\frac35\right)+\frac45
\displaystyle =-\frac32+3+\frac45
\displaystyle =\frac{-15+30+8}{10}=\frac{23}{10}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{The value of }\sin^2 75^\circ-\sin^2 15^\circ\text{ is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }\frac{\sqrt3}{2}\qquad\text{(c) }1\qquad\text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sin75^\circ=\cos15^\circ.
\displaystyle \therefore \sin^2 75^\circ-\sin^2 15^\circ
\displaystyle =\cos^2 15^\circ-\sin^2 15^\circ
\displaystyle =\cos30^\circ=\frac{\sqrt3}{2}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{If }A+B+C=180^\circ,\text{ then }  \sec A(\cos B\cos C-\sin B\sin C)\text{ is equal to}
\displaystyle \text{(a) }0\qquad\text{(b) }-1\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos B\cos C-\sin B\sin C=\cos(B+C).
\displaystyle A+B+C=180^\circ\Rightarrow B+C=180^\circ-A.
\displaystyle \therefore \cos(B+C)=\cos(180^\circ-A)=-\cos A.
\displaystyle \therefore \sec A(\cos B\cos C-\sin B\sin C)
\displaystyle =\sec A(-\cos A)=-1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ\text{ is equal to}
\displaystyle \text{(a) }\frac{\sqrt3}{4}\qquad\text{(b) }\frac{\sqrt3}{2}\qquad\text{(c) }\sqrt3\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \tan(20^\circ+40^\circ)=\tan60^\circ=\sqrt3.
\displaystyle \frac{\tan20^\circ+\tan40^\circ}{1-\tan20^\circ\tan40^\circ}=\sqrt3.
\displaystyle \therefore \tan20^\circ+\tan40^\circ
\displaystyle =\sqrt3\{1-\tan20^\circ\tan40^\circ\}.
\displaystyle \therefore \tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ
\displaystyle =\sqrt3.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If }\tan A=\frac{a}{a+1}\text{ and }\tan B=\frac{1}{2a+1},  \text{ then the value of } \\ A+B\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }\frac{\pi}{2}\qquad\text{(c) }\frac{\pi}{3}\qquad\text{(d) }\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.
\displaystyle =\frac{\frac{a}{a+1}+\frac{1}{2a+1}}{1-\frac{a}{(a+1)(2a+1)}}.
\displaystyle =\frac{2a^2+2a+1}{2a^2+2a+1}=1.
\displaystyle \therefore \tan(A+B)=1.
\displaystyle \therefore A+B=\frac{\pi}{4}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{If }3\sin\theta+4\cos\theta=5,\text{ then }4\sin\theta-3\cos\theta=
\displaystyle \text{(a) }0\qquad\text{(b) }5\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 3\sin\theta+4\cos\theta\leq\sqrt{3^2+4^2}=5.
\displaystyle \text{Since the maximum value }5\text{ is attained,}
\displaystyle \sin\theta=\frac35,\qquad\cos\theta=\frac45.
\displaystyle \therefore 4\sin\theta-3\cos\theta
\displaystyle =4\left(\frac35\right)-3\left(\frac45\right)=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{If in a }\triangle ABC,\ \tan A+\tan B+\tan C=6,\text{ then}
\displaystyle \cot A\cot B\cot C=
\displaystyle \text{(a) }6\qquad\text{(b) }1\qquad\text{(c) }\frac16\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Since }A+B+C=\pi,
\displaystyle \tan A+\tan B+\tan C=\tan A\tan B\tan C.
\displaystyle \therefore \tan A\tan B\tan C=6.
\displaystyle \therefore \cot A\cot B\cot C
\displaystyle =\frac{1}{\tan A\tan B\tan C}=\frac16.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\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 \tan3A=\tan(2A+A)
\displaystyle =\frac{\tan2A+\tan A}{1-\tan2A\tan A}.
\displaystyle \therefore \tan3A(1-\tan2A\tan A)=\tan2A+\tan A.
\displaystyle \therefore \tan3A-\tan2A-\tan A
\displaystyle =\tan3A\tan2A\tan A.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{If }A+B+C=180^\circ,\text{ then } \frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}\text{ is equal to}
\displaystyle \text{(a) }\tan A\tan B\tan C\qquad\text{(b) }0\qquad\text{(c) }1
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle A+B+C=180^\circ=\pi.
\displaystyle \therefore \tan(A+B+C)=\tan\pi=0.
\displaystyle \therefore \tan A+\tan B+\tan C-\tan A\tan B\tan C=0.
\displaystyle \therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C.
\displaystyle \therefore \frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}=1.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{If }\cos P=\frac17\text{ and }\cos Q=\frac{13}{14},\text{ where }P\text{ and }Q
\displaystyle \text{both are acute angles. Then, the value of }P-Q\text{ is}
\displaystyle \text{(a) }30^\circ\qquad\text{(b) }60^\circ\qquad\text{(c) }45^\circ\qquad\text{(d) }75^\circ
\displaystyle \text{Answer:}
\displaystyle \cos P=\frac17\Rightarrow\sin P=\sqrt{1-\frac1{49}}=\frac{4\sqrt3}{7}.
\displaystyle \cos Q=\frac{13}{14}\Rightarrow\sin Q=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt3}{14}.
\displaystyle \cos(P-Q)=\cos P\cos Q+\sin P\sin Q
\displaystyle =\frac17\cdot\frac{13}{14}+\frac{4\sqrt3}{7}\cdot\frac{3\sqrt3}{14}
\displaystyle =\frac{13}{98}+\frac{36}{98}=\frac12.
\displaystyle \text{Since }P,Q\text{ are acute and }P>Q,\quad0<P-Q<90^\circ.
\displaystyle \therefore P-Q=60^\circ.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }\cot(\alpha+\beta)=0,\text{ then }\sin(\alpha+2\beta)\text{ is equal to}
\displaystyle \text{(a) }\sin\alpha\qquad\text{(b) }\cos2\beta\qquad\text{(c) }\cos\alpha\qquad\text{(d) }\sin2\alpha
\displaystyle \text{Answer:}
\displaystyle \cot(\alpha+\beta)=0.
\displaystyle \text{Taking }\alpha+\beta=\frac{\pi}{2},
\displaystyle \alpha+2\beta=(\alpha+\beta)+\beta=\frac{\pi}{2}+\beta.
\displaystyle \therefore \sin(\alpha+2\beta)=\sin\left(\frac{\pi}{2}+\beta\right)=\cos\beta.
\displaystyle \text{Also, }\alpha+\beta=\frac{\pi}{2}\Rightarrow\beta=\frac{\pi}{2}-\alpha.
\displaystyle \therefore \cos\beta=\cos\left(\frac{\pi}{2}-\alpha\right)=\sin\alpha.
\displaystyle \therefore \sin(\alpha+2\beta)=\sin\alpha.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}\text{ is equal to}
\displaystyle \text{(a) }\tan55^\circ\qquad\text{(b) }\cot55^\circ\qquad\text{(c) }-\tan35^\circ\qquad\text{(d) }-\cot35^\circ
\displaystyle \text{Answer:}
\displaystyle \frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}
\displaystyle =\frac{1+\tan10^\circ}{1-\tan10^\circ}
\displaystyle =\tan(45^\circ+10^\circ)
\displaystyle =\tan55^\circ.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{The value of }\cos^2\left(\frac{\pi}{6}+\theta\right)   -\sin^2\left(\frac{\pi}{6}-\theta\right)\text{ is}
\displaystyle \text{(a) }\frac12\cos2\theta\qquad\text{(b) }0\qquad\text{(c) }-\frac12\cos2\theta\qquad\text{(d) }\frac12
\displaystyle \text{Answer:}
\displaystyle \text{Using }\cos^2A-\sin^2B=\frac12(\cos2A+\cos2B),
\displaystyle \cos^2\left(\frac{\pi}{6}+\theta\right)-\sin^2\left(\frac{\pi}{6}-\theta\right)
\displaystyle =\frac12\left\{\cos\left(\frac{\pi}{3}+2\theta\right)+\cos\left(\frac{\pi}{3}-2\theta\right)\right\}
\displaystyle =\frac12\left\{2\cos\frac{\pi}{3}\cos2\theta\right\}
\displaystyle =\frac12\cos2\theta.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{If }\tan\theta_1\tan\theta_2=k,\text{ then }\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=
\displaystyle \text{(a) }\frac{1+k}{1-k}\qquad\text{(b) }\frac{1-k}{1+k}\qquad\text{(c) }\frac{k+1}{k-1}\qquad\text{(d) }\frac{k-1}{k+1}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}
\displaystyle =\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2} {\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2}
\displaystyle =\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}
\displaystyle =\frac{1+k}{1-k}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{If }\sin(\pi\cos\theta)=\cos(\pi\sin\theta),\text{ then }\sin2\theta=
\displaystyle \text{(a) }\pm\frac34\qquad\text{(b) }\pm\frac43\qquad\text{(c) }\pm\frac13\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos(\pi\sin\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right).
\displaystyle \therefore \sin(\pi\cos\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right).
\displaystyle \text{Hence either }\cos\theta+\sin\theta=\frac12
\displaystyle \text{or }\cos\theta-\sin\theta=\frac12.
\displaystyle \text{If }\cos\theta+\sin\theta=\frac12,\text{ then}
\displaystyle (\cos\theta+\sin\theta)^2=1+\sin2\theta=\frac14.
\displaystyle \therefore \sin2\theta=-\frac34.
\displaystyle \text{If }\cos\theta-\sin\theta=\frac12,\text{ then}
\displaystyle (\cos\theta-\sin\theta)^2=1-\sin2\theta=\frac14.
\displaystyle \therefore \sin2\theta=\frac34.
\displaystyle \therefore \sin2\theta=\pm\frac34.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{If }\tan\theta=\frac12\text{ and }\tan\phi=\frac13,\text{ then the value of }\theta+\phi\text{ is}
\displaystyle \text{(a) }\frac{\pi}{6}\qquad\text{(b) }\pi\qquad\text{(c) }0\qquad\text{(d) }\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \tan(\theta+\phi)=\frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}
\displaystyle =\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}
\displaystyle =\frac{\frac56}{\frac56}=1.
\displaystyle \therefore \theta+\phi=\frac{\pi}{4}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{The value of } \\ \cos(36^\circ-A)\cos(36^\circ+A)  +\cos(54^\circ+A)\cos(54^\circ-A)\text{ is}
\displaystyle \text{(a) }\sin2A\qquad\text{(b) }\cos2A\qquad\text{(c) }\cos3A\qquad\text{(d) }\sin3A
\displaystyle \text{Answer:}
\displaystyle \cos(36^\circ-A)\cos(36^\circ+A)
\displaystyle =\frac12\{\cos2A+\cos72^\circ\}.
\displaystyle \cos(54^\circ+A)\cos(54^\circ-A)
\displaystyle =\frac12\{\cos2A+\cos108^\circ\}.
\displaystyle \therefore \text{Required expression}
\displaystyle =\cos2A+\frac12(\cos72^\circ+\cos108^\circ).
\displaystyle \cos108^\circ=\cos(180^\circ-72^\circ)=-\cos72^\circ.
\displaystyle \therefore \text{Required expression}=\cos2A.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{If }\tan\left(\frac{\pi}{4}+\theta\right)+\tan\left(\frac{\pi}{4}-\theta\right)=a,\text{ then}
\displaystyle \tan^2\left(\frac{\pi}{4}+\theta\right)+\tan^2\left(\frac{\pi}{4}-\theta\right)=
\displaystyle \text{(a) }a^2+1\qquad\text{(b) }a^2+2\qquad\text{(c) }a^2-2\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x=\tan\left(\frac{\pi}{4}+\theta\right),\quad y=\tan\left(\frac{\pi}{4}-\theta\right).
\displaystyle \text{Then }x+y=a.
\displaystyle xy=\tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{\pi}{4}-\theta\right)=1.
\displaystyle \therefore x^2+y^2=(x+y)^2-2xy
\displaystyle =a^2-2.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{If }\tan(A-B)=1,\ \sec(A+B)=\frac{2}{\sqrt3},\text{ then}
\displaystyle \text{the smallest positive value of }B\text{ is}
\displaystyle \text{(a) }\frac{25\pi}{24}\qquad\text{(b) }\frac{19\pi}{24}\qquad\text{(c) }\frac{13\pi}{24}\qquad\text{(d) }\frac{11\pi}{24}
\displaystyle \text{Answer:}
\displaystyle \tan(A-B)=1
\displaystyle \Rightarrow A-B=\frac{\pi}{4}.
\displaystyle \sec(A+B)=\frac{2}{\sqrt3}
\displaystyle \Rightarrow \cos(A+B)=\frac{\sqrt3}{2}.
\displaystyle \therefore A+B=\frac{11\pi}{6}.
\displaystyle \text{Subtracting }A-B=\frac{\pi}{4}\text{ from }A+B=\frac{11\pi}{6},
\displaystyle 2B=\frac{11\pi}{6}-\frac{\pi}{4}
\displaystyle =\frac{22\pi-3\pi}{12}=\frac{19\pi}{12}.
\displaystyle \therefore B=\frac{19\pi}{24}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{If }A-B=\frac{\pi}{4},\text{ then }(1+\tan A)(1-\tan B)\text{ is equal to}
\displaystyle \text{(a) }2\qquad\text{(b) }1\qquad\text{(c) }0\qquad\text{(d) }3
\displaystyle \text{Answer:}
\displaystyle A-B=\frac{\pi}{4}\Rightarrow\tan(A-B)=1.
\displaystyle \therefore \frac{\tan A-\tan B}{1+\tan A\tan B}=1.
\displaystyle \therefore \tan A-\tan B=1+\tan A\tan B.
\displaystyle (1+\tan A)(1-\tan B)
\displaystyle =1+\tan A-\tan B-\tan A\tan B
\displaystyle =1+(1+\tan A\tan B)-\tan A\tan B=2.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{The maximum value of }\sin^2(120^\circ+\theta)+\sin^2(120^\circ-\theta)\text{ is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }\frac32\qquad\text{(c) }\frac54\qquad\text{(d) }\frac34
\displaystyle \text{Answer:}
\displaystyle \sin^2x=\frac{1-\cos2x}{2}.
\displaystyle \therefore \sin^2(120^\circ+\theta)+\sin^2(120^\circ-\theta)
\displaystyle =1-\frac12\{\cos(240^\circ+2\theta)+\cos(240^\circ-2\theta)\}
\displaystyle =1-\cos240^\circ\cos2\theta.
\displaystyle \text{Since }\cos240^\circ=-\frac12,
\displaystyle \text{the expression}=1+\frac12\cos2\theta.
\displaystyle \text{Since }\cos2\theta\leq1,\text{ maximum value}=1+\frac12=\frac32.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{If }\cos(A-B)=\frac35\text{ and }\tan A\tan B=2,\text{ then}
\displaystyle \text{(a) }\cos A\cos B=\frac15\qquad\text{(b) }\cos A\cos B=-\frac15
\displaystyle \text{(c) }\sin A\sin B=\frac15\qquad\text{(d) }\sin A\sin B=-\frac15
\displaystyle \text{Answer:}
\displaystyle \tan A\tan B=2.
\displaystyle \therefore \frac{\sin A\sin B}{\cos A\cos B}=2
\displaystyle \Rightarrow \sin A\sin B=2\cos A\cos B.
\displaystyle \cos(A-B)=\cos A\cos B+\sin A\sin B
\displaystyle =3\cos A\cos B.
\displaystyle \therefore 3\cos A\cos B=\frac35
\displaystyle \Rightarrow \cos A\cos B=\frac15.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{If }\tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=2k,\text{ then }k=
\displaystyle \text{(a) }-1\qquad\text{(b) }\frac12\qquad\text{(c) }-\frac12\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 69^\circ+66^\circ=135^\circ.
\displaystyle \tan135^\circ=-1.
\displaystyle \frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}=-1.
\displaystyle \therefore \tan69^\circ+\tan66^\circ
\displaystyle =-1+\tan69^\circ\tan66^\circ.
\displaystyle \therefore \tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=-1.
\displaystyle \therefore 2k=-1\Rightarrow k=-\frac12.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{If }\tan\alpha=\frac{x}{x+1}\text{ and }\tan\beta=\frac1{2x+1},   \text{ then }\alpha+\beta\text{ is equal to}
\displaystyle \text{(a) }\frac{\pi}{2}\qquad\text{(b) }\frac{\pi}{3}\qquad\text{(c) }\frac{\pi}{6}\qquad\text{(d) }\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
\displaystyle =\frac{\frac{x}{x+1}+\frac1{2x+1}}{1-\frac{x}{(x+1)(2x+1)}}.
\displaystyle =\frac{2x^2+2x+1}{2x^2+2x+1}=1.
\displaystyle \therefore \tan(\alpha+\beta)=1.
\displaystyle \therefore \alpha+\beta=\frac{\pi}{4}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{Write the maximum and minimum values of }\cos(\cos x).
\displaystyle \text{Answer:}
\displaystyle -1\leq\cos x\leq1.
\displaystyle \text{Let }t=\cos x.\text{ Then }-1\leq t\leq1.
\displaystyle \text{On }[-1,1],\text{ the maximum value of }\cos t\text{ is }\cos0=1.
\displaystyle \text{The minimum value is attained at }t=\pm1.
\displaystyle \therefore \text{Maximum value}=1\text{ and minimum value}=\cos1.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the maximum and minimum values of }\sin(\sin x).
\displaystyle \text{Answer:}
\displaystyle -1\leq\sin x\leq1.
\displaystyle \text{Let }t=\sin x.\text{ Then }-1\leq t\leq1.
\displaystyle \sin t\text{ is increasing on }[-1,1].
\displaystyle \therefore \text{Maximum value}=\sin1.
\displaystyle \text{Minimum value}=\sin(-1)=-\sin1.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Write the maximum value of }\sin(\cos x).
\displaystyle \text{Answer:}
\displaystyle -1\leq\cos x\leq1.
\displaystyle \text{Let }t=\cos x.\text{ Then }-1\leq t\leq1.
\displaystyle \sin t\text{ is increasing on }[-1,1].
\displaystyle \therefore \text{Maximum value of }\sin(\cos x)=\sin1.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\sin x=\cos^2x,\text{ then write the value of}
\displaystyle \cos^2x(1+\cos^2x).
\displaystyle \text{Answer:}
\displaystyle \sin x=\cos^2x=1-\sin^2x.
\displaystyle \therefore \sin^2x+\sin x=1.
\displaystyle \cos^2x(1+\cos^2x)=\sin x(1+\sin x)
\displaystyle =\sin x+\sin^2x=1.
\displaystyle \therefore \text{Required value}=1.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{If }\sin x+\mathrm{cosec}\,x=2,\text{ then write the value of}
\displaystyle \sin^n x+\mathrm{cosec}^n x.
\displaystyle \text{Answer:}
\displaystyle \sin x+\frac1{\sin x}=2.
\displaystyle \text{Let }t=\sin x,\quad t\ne0.
\displaystyle t+\frac1t=2
\displaystyle \Rightarrow t^2-2t+1=0
\displaystyle \Rightarrow (t-1)^2=0\Rightarrow t=1.
\displaystyle \therefore \sin x=1\text{ and }\mathrm{cosec}\,x=1.
\displaystyle \therefore \sin^n x+\mathrm{cosec}^n x=1^n+1^n=2.
\displaystyle \therefore \text{Required value}=2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\sin x+\sin^2x=1,\text{ then write the value of}
\displaystyle \cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x.
\displaystyle \text{Answer:}
\displaystyle \sin x+\sin^2x=1.
\displaystyle \therefore \sin x=1-\sin^2x=\cos^2x.
\displaystyle \cos^{12}x+3\cos^{10}x+3\cos^8x+\cos^6x
\displaystyle =\cos^6x(\cos^6x+3\cos^4x+3\cos^2x+1)
\displaystyle =\cos^6x(1+\cos^2x)^3
\displaystyle =\{\cos^2x(1+\cos^2x)\}^3.
\displaystyle \text{But }\cos^2x(1+\cos^2x)=\sin x(1+\sin x)
\displaystyle =\sin x+\sin^2x=1.
\displaystyle \therefore \text{Required value}=1^3=1.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If }\sin x+\sin^2x=1,\text{ then write the value of}
\displaystyle \cos^8x+2\cos^6x+\cos^4x.
\displaystyle \text{Answer:}
\displaystyle \sin x+\sin^2x=1.
\displaystyle \therefore \sin x=1-\sin^2x=\cos^2x.
\displaystyle \cos^8x+2\cos^6x+\cos^4x
\displaystyle =\cos^4x(\cos^4x+2\cos^2x+1)
\displaystyle =\cos^4x(1+\cos^2x)^2
\displaystyle =\{\cos^2x(1+\cos^2x)\}^2.
\displaystyle =\{\sin x(1+\sin x)\}^2
\displaystyle =(\sin x+\sin^2x)^2=1.
\displaystyle \therefore \text{Required value}=1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\sin\theta_1+\sin\theta_2+\sin\theta_3=3,\text{ then write the value of}
\displaystyle \cos\theta_1+\cos\theta_2+\cos\theta_3.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\sin\theta_i\leq1\text{ for }i=1,2,3,
\displaystyle \sin\theta_1+\sin\theta_2+\sin\theta_3\leq3.
\displaystyle \text{Given that the sum is }3,\text{ each term must be }1.
\displaystyle \therefore \sin\theta_1=\sin\theta_2=\sin\theta_3=1.
\displaystyle \therefore \cos\theta_1=\cos\theta_2=\cos\theta_3=0.
\displaystyle \therefore \cos\theta_1+\cos\theta_2+\cos\theta_3=0.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the value of }\sin10^\circ+\sin20^\circ+\sin30^\circ+\cdots+\sin360^\circ.
\displaystyle \text{Answer:}
\displaystyle \sin(360^\circ-\theta)=-\sin\theta.
\displaystyle \therefore \sin10^\circ+\sin350^\circ=0,
\displaystyle \sin20^\circ+\sin340^\circ=0,\quad\ldots,\quad\sin170^\circ+\sin190^\circ=0.
\displaystyle \text{Also, }\sin180^\circ=0\text{ and }\sin360^\circ=0.
\displaystyle \therefore \sin10^\circ+\sin20^\circ+\sin30^\circ+\cdots+\sin360^\circ=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{A circular wire of radius }15\text{ cm is cut and bent so as to lie}
\displaystyle \text{along the circumference of a loop of radius }120\text{ cm. Write the measure}
\displaystyle \text{of the angle subtended by it at the centre of the loop.}
\displaystyle \text{Answer:}
\displaystyle \text{Length of the circular wire}=2\pi(15)=30\pi\text{ cm}.
\displaystyle \text{This becomes the length of the arc of the loop.}
\displaystyle \text{Using }s=r\theta,
\displaystyle 30\pi=120\theta
\displaystyle \Rightarrow \theta=\frac{30\pi}{120}=\frac{\pi}{4}.
\displaystyle \therefore \text{Required angle}=\frac{\pi}{4}\text{ radians}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Write the value of }2(\sin^6\theta+\cos^6\theta)
\displaystyle -3(\sin^4\theta+\cos^4\theta)+1.
\displaystyle \text{Answer:}
\displaystyle \text{Let }a=\sin^2\theta,\quad b=\cos^2\theta.
\displaystyle \therefore a+b=1.
\displaystyle a^3+b^3=(a+b)^3-3ab(a+b)=1-3ab.
\displaystyle a^2+b^2=(a+b)^2-2ab=1-2ab.
\displaystyle \therefore 2(a^3+b^3)-3(a^2+b^2)+1
\displaystyle =2(1-3ab)-3(1-2ab)+1
\displaystyle =2-6ab-3+6ab+1=0.
\displaystyle \therefore \text{Required value}=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Write the value of }\cos1^\circ+\cos2^\circ+\cos3^\circ+\cdots+\cos180^\circ.
\displaystyle \text{Answer:}
\displaystyle \cos(180^\circ-\theta)=-\cos\theta.
\displaystyle \therefore \cos1^\circ+\cos179^\circ=0,
\displaystyle \cos2^\circ+\cos178^\circ=0,\quad\ldots,\quad\cos89^\circ+\cos91^\circ=0.
\displaystyle \text{Also, }\cos90^\circ=0\text{ and }\cos180^\circ=-1.
\displaystyle \therefore \cos1^\circ+\cos2^\circ+\cdots+\cos180^\circ=-1.
\displaystyle \therefore \text{Required value}=-1.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\cot(\alpha+\beta)=0,\text{ then write the value of }\sin(\alpha+2\beta).
\displaystyle \text{Answer:}
\displaystyle \cot(\alpha+\beta)=0\Rightarrow\cos(\alpha+\beta)=0.
\displaystyle \therefore \alpha+\beta=\frac{(2n+1)\pi}{2},\quad n\in Z.
\displaystyle \therefore \sin(\alpha+\beta)=(-1)^n.
\displaystyle \sin(\alpha+2\beta)=\sin\{(\alpha+\beta)+\beta\}
\displaystyle =\sin(\alpha+\beta)\cos\beta+\cos(\alpha+\beta)\sin\beta
\displaystyle =(-1)^n\cos\beta.
\displaystyle \therefore \sin(\alpha+2\beta)=\pm\cos\beta.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\tan A+\cot A=4,\text{ then write the value of}
\displaystyle \tan^4A+\cot^4A.
\displaystyle \text{Answer:}
\displaystyle \tan A+\cot A=4.
\displaystyle \text{Squaring both sides,}
\displaystyle \tan^2A+\cot^2A+2\tan A\cot A=16.
\displaystyle \text{Since }\tan A\cot A=1,
\displaystyle \tan^2A+\cot^2A=14.
\displaystyle \text{Squaring again,}
\displaystyle \tan^4A+\cot^4A+2\tan^2A\cot^2A=196.
\displaystyle \therefore \tan^4A+\cot^4A=196-2=194.
\displaystyle \therefore \text{Required value}=194.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Write the least value of }\cos^2\theta+\sec^2\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\cos^2\theta.
\displaystyle \text{Since }\sec\theta\text{ is defined, }t>0.
\displaystyle \cos^2\theta+\sec^2\theta=t+\frac1t.
\displaystyle \text{For }t>0,\quad t+\frac1t\geq2.
\displaystyle \text{Equality holds when }t=1.
\displaystyle \therefore \text{Least value}=2.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }x=\sin^{14}\theta+\cos^{20}\theta,\text{ then write the smallest}
\displaystyle \text{interval in which the value of }x\text{ lie.}
\displaystyle \text{Answer:}
\displaystyle 0\leq\sin^2\theta\leq1\text{ and }0\leq\cos^2\theta\leq1.
\displaystyle \therefore \sin^{14}\theta\leq\sin^2\theta
\displaystyle \text{and }\cos^{20}\theta\leq\cos^2\theta.
\displaystyle \therefore x=\sin^{14}\theta+\cos^{20}\theta
\displaystyle \leq\sin^2\theta+\cos^2\theta=1.
\displaystyle \text{Also, }\sin^{14}\theta\text{ and }\cos^{20}\theta\text{ cannot both be zero.}
\displaystyle \therefore x>0.
\displaystyle \therefore 0<x\leq1.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{If }3\sin\theta+5\cos\theta=5,\text{ then write the value of}
\displaystyle 5\sin\theta-3\cos\theta.
\displaystyle \text{Answer:}
\displaystyle (3\sin\theta+5\cos\theta)^2+(5\sin\theta-3\cos\theta)^2
\displaystyle =34(\sin^2\theta+\cos^2\theta)=34.
\displaystyle \text{Given, }3\sin\theta+5\cos\theta=5.
\displaystyle \therefore 25+(5\sin\theta-3\cos\theta)^2=34.
\displaystyle \Rightarrow (5\sin\theta-3\cos\theta)^2=9.
\displaystyle \therefore 5\sin\theta-3\cos\theta=\pm3.
\displaystyle \therefore \text{Required value}=\pm3.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{If }\alpha+\beta-\gamma=\pi,\text{ and}
\displaystyle \sin^2\alpha+\sin^2\beta-\sin^2\gamma=\lambda\sin\alpha\sin\beta\cos\gamma,
\displaystyle \text{then write the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta-\gamma=\pi\Rightarrow\alpha+\beta=\pi+\gamma.
\displaystyle \therefore \sin(\alpha+\beta)=\sin(\pi+\gamma)=-\sin\gamma.
\displaystyle \sin^2\gamma=\sin^2(\alpha+\beta).
\displaystyle \therefore \sin^2\alpha+\sin^2\beta-\sin^2\gamma
\displaystyle =\sin^2\alpha+\sin^2\beta-\sin^2(\alpha+\beta).
\displaystyle =\sin^2\alpha+\sin^2\beta-(\sin\alpha\cos\beta+\cos\alpha\sin\beta)^2
\displaystyle =2\sin\alpha\sin\beta(\sin\alpha\sin\beta-\cos\alpha\cos\beta)
\displaystyle =-2\sin\alpha\sin\beta\cos(\alpha+\beta).
\displaystyle \text{Since }\cos(\alpha+\beta)=\cos(\pi+\gamma)=-\cos\gamma,
\displaystyle \sin^2\alpha+\sin^2\beta-\sin^2\gamma=2\sin\alpha\sin\beta\cos\gamma.
\displaystyle \therefore \lambda=2.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }x\cos\theta=y\cos\left(\theta+\frac{2\pi}{3}\right)
\displaystyle =z\cos\left(\theta+\frac{4\pi}{3}\right),\text{ then write the value of }\frac1x+\frac1y+\frac1z.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\cos\theta=y\cos\left(\theta+\frac{2\pi}{3}\right)
\displaystyle =z\cos\left(\theta+\frac{4\pi}{3}\right)=k.
\displaystyle \therefore \frac1x=\frac{\cos\theta}{k},\quad
\displaystyle \frac1y=\frac{\cos(\theta+2\pi/3)}{k},\quad
\displaystyle \frac1z=\frac{\cos(\theta+4\pi/3)}{k}.
\displaystyle \therefore \frac1x+\frac1y+\frac1z
\displaystyle =\frac1k\left\{\cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right) +\cos\left(\theta+\frac{4\pi}{3}\right)\right\}.
\displaystyle \cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right)  +\cos\left(\theta+\frac{4\pi}{3}\right)=0.
\displaystyle \therefore \frac1x+\frac1y+\frac1z=0.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the maximum and minimum values of }
\displaystyle 3\cos x+4\sin x+5.
\displaystyle \text{Answer:}
\displaystyle \text{For }a\cos x+b\sin x,\text{ maximum value}=\sqrt{a^2+b^2}
\displaystyle \text{and minimum value}=-\sqrt{a^2+b^2}.
\displaystyle \therefore -5\leq3\cos x+4\sin x\leq5.
\displaystyle \text{Adding }5\text{ throughout,}
\displaystyle 0\leq3\cos x+4\sin x+5\leq10.
\displaystyle \therefore \text{Maximum value}=10\text{ and minimum value}=0.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Write the maximum value of }12\sin\theta-9\sin^2\theta.
\displaystyle \text{Answer:}
\displaystyle 12\sin\theta-9\sin^2\theta
\displaystyle =-9\left(\sin^2\theta-\frac43\sin\theta\right)
\displaystyle =-9\left\{\left(\sin\theta-\frac23\right)^2-\frac49\right\}
\displaystyle =4-9\left(\sin\theta-\frac23\right)^2.
\displaystyle \text{Since }\left(\sin\theta-\frac23\right)^2\geq0,
\displaystyle 12\sin\theta-9\sin^2\theta\leq4.
\displaystyle \text{Equality occurs when }\sin\theta=\frac23.
\displaystyle \therefore \text{Maximum value}=4.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }12\sin\theta-9\sin^2\theta\text{ attains its maximum value}
\displaystyle \text{at }\theta=\alpha,\text{ then write the value of }\sin\alpha.
\displaystyle \text{Answer:}
\displaystyle 12\sin\theta-9\sin^2\theta
\displaystyle =4-9\left(\sin\theta-\frac23\right)^2.
\displaystyle \text{The maximum value is attained when}
\displaystyle \left(\sin\theta-\frac23\right)^2=0.
\displaystyle \therefore \sin\theta=\frac23.
\displaystyle \text{Since the maximum occurs at }\theta=\alpha,
\displaystyle \therefore \sin\alpha=\frac23.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Write the interval in which the values of}
\displaystyle 5\cos\theta+3\cos\left(\theta+\frac{\pi}{3}\right)+3\text{ lie.}
\displaystyle \text{Answer:}
\displaystyle 5\cos\theta+3\cos\left(\theta+\frac{\pi}{3}\right)+3
\displaystyle =5\cos\theta+3\left(\cos\theta\cos\frac{\pi}{3} -\sin\theta\sin\frac{\pi}{3}\right)+3
\displaystyle =\frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta+3.
\displaystyle \text{For }a\cos\theta+b\sin\theta,\text{ maximum magnitude}=\sqrt{a^2+b^2}.
\displaystyle \sqrt{\left(\frac{13}{2}\right)^2+\left(-\frac{3\sqrt3}{2}\right)^2}
\displaystyle =\frac12\sqrt{169+27}=\frac{14}{2}=7.
\displaystyle \therefore -7\leq\frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta\leq7.
\displaystyle \text{Adding }3\text{ throughout,}
\displaystyle -4\leq5\cos\theta+3\cos\left(\theta+\frac{\pi}{3}\right)+3\leq10.
\displaystyle \therefore \text{Required interval}=[-4,10].
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{If }\tan(A+B)=p\text{ and }\tan(A-B)=q,\text{ then write the value of }\tan2B.
\displaystyle \text{Answer:}
\displaystyle 2B=(A+B)-(A-B).
\displaystyle \therefore \tan2B=\tan\{(A+B)-(A-B)\}
\displaystyle =\frac{\tan(A+B)-\tan(A-B)}{1+\tan(A+B)\tan(A-B)}.
\displaystyle =\frac{p-q}{1+pq}.
\displaystyle \therefore \tan2B=\frac{p-q}{1+pq}.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\frac{\cos(x-y)}{\cos(x+y)}=\frac{m}{n},\text{ then write the value of }\tan x\tan y.
\displaystyle \text{Answer:}
\displaystyle \frac{\cos x\cos y+\sin x\sin y}{\cos x\cos y-\sin x\sin y}=\frac{m}{n}.
\displaystyle \therefore \frac{1+\tan x\tan y}{1-\tan x\tan y}=\frac{m}{n}.
\displaystyle \text{Let }t=\tan x\tan y.
\displaystyle \therefore n(1+t)=m(1-t).
\displaystyle \Rightarrow n+nt=m-mt
\displaystyle \Rightarrow (m+n)t=m-n.
\displaystyle \therefore t=\frac{m-n}{m+n}.
\displaystyle \therefore \tan x\tan y=\frac{m-n}{m+n}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{If }a=b\cos\frac{2\pi}{3}=c\cos\frac{4\pi}{3},\text{ then write the value of }ab+bc+ca.
\displaystyle \text{Answer:}
\displaystyle \cos\frac{2\pi}{3}=-\frac12\text{ and }\cos\frac{4\pi}{3}=-\frac12.
\displaystyle \therefore a=-\frac b2=-\frac c2.
\displaystyle \therefore b=-2a\text{ and }c=-2a.
\displaystyle ab+bc+ca=a(-2a)+(-2a)(-2a)+(-2a)a
\displaystyle =-2a^2+4a^2-2a^2=0.
\displaystyle \therefore ab+bc+ca=0.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }A+B=C,\text{ then write the value of }\tan A\tan B\tan C.
\displaystyle \text{Answer:}
\displaystyle A+B=C.
\displaystyle \therefore \tan C=\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}.
\displaystyle \tan C(1-\tan A\tan B)=\tan A+\tan B.
\displaystyle \therefore \tan A\tan B\tan C=\tan C-\tan A-\tan B.
\displaystyle \therefore \text{Required value}=\tan C-\tan A-\tan B.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{If }\sin\alpha-\sin\beta=a\text{ and }\cos\alpha+\cos\beta=b,
\displaystyle \text{then write the value of }\cos(\alpha+\beta).
\displaystyle \text{Answer:}
\displaystyle \sin\alpha-\sin\beta=a,\qquad\cos\alpha+\cos\beta=b.
\displaystyle \text{Squaring and adding,}
\displaystyle a^2+b^2=(\sin\alpha-\sin\beta)^2+(\cos\alpha+\cos\beta)^2
\displaystyle =\sin^2\alpha+\cos^2\alpha+\sin^2\beta+\cos^2\beta
\displaystyle \quad+2(\cos\alpha\cos\beta-\sin\alpha\sin\beta)
\displaystyle =2+2\cos(\alpha+\beta).
\displaystyle \therefore \cos(\alpha+\beta)=\frac{a^2+b^2-2}{2}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }\tan\alpha=\frac{1}{1+2^{-x}}\text{ and }\tan\beta=\frac{1}{1+2^{x+1}},
\displaystyle \text{then write the value of }\alpha+\beta\text{ lying in the interval }\left(0,\frac{\pi}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=2^x.
\displaystyle \therefore \tan\alpha=\frac{t}{t+1},\qquad\tan\beta=\frac{1}{1+2t}.
\displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
\displaystyle =\frac{\frac{t}{t+1}+\frac{1}{1+2t}}{1-\frac{t}{(t+1)(1+2t)}}.
\displaystyle =\frac{2t^2+2t+1}{2t^2+2t+1}=1.
\displaystyle \therefore \tan(\alpha+\beta)=1.
\displaystyle \text{Since }0<\alpha+\beta<\frac{\pi}{2},
\displaystyle \therefore \alpha+\beta=\frac{\pi}{4}.
\displaystyle \\

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