\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 \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 \\

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