\displaystyle \textbf{Question 1. }8\sin\frac{x}{8}\cos\frac{x}{2}\cos\frac{x}{4}\cos\frac{x}{8}\text{ is equal to}
\displaystyle \text{(a) }8\cos x\qquad \text{(b) }\cos x\qquad \text{(c) }8\sin x\qquad \text{(d) }\sin x
\displaystyle \text{Answer:}
\displaystyle 2\sin\frac{x}{8}\cos\frac{x}{8}=\sin\frac{x}{4}
\displaystyle 4\sin\frac{x}{8}\cos\frac{x}{8}\cos\frac{x}{4}=2\sin\frac{x}{4}\cos\frac{x}{4}=\sin\frac{x}{2}
\displaystyle 8\sin\frac{x}{8}\cos\frac{x}{8}\cos\frac{x}{4}\cos\frac{x}{2}=2\sin\frac{x}{2}\cos\frac{x}{2}=\sin x
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 2. }\frac{\sec8A-1}{\sec4A-1}\text{ is equal to}
\displaystyle \text{(a) }\frac{\tan2A}{\tan8A}\qquad \text{(b) }\frac{\tan8A}{\tan2A}\qquad \text{(c) }\frac{\cot8A}{\cot2A}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{\sec8A-1}{\sec4A-1}=\frac{\frac{1}{\cos8A}-1}{\frac{1}{\cos4A}-1}
\displaystyle =\frac{\frac{1-\cos8A}{\cos8A}}{\frac{1-\cos4A}{\cos4A}}
\displaystyle =\frac{1-\cos8A}{1-\cos4A}\cdot\frac{\cos4A}{\cos8A}
\displaystyle =\frac{2\sin^24A}{2\sin^22A}\cdot\frac{\cos4A}{\cos8A}
\displaystyle =\frac{\sin^24A}{\sin^22A}\cdot\frac{\cos4A}{\cos8A}
\displaystyle =\frac{(2\sin2A\cos2A)^2}{\sin^22A}\cdot\frac{\cos4A}{\cos8A}
\displaystyle =4\cos^22A\cdot\frac{\cos4A}{\cos8A}
\displaystyle =\frac{2\sin4A\cos4A}{2\sin2A\cos2A}\cdot\frac{\cos2A}{\cos8A}
\displaystyle =\frac{\sin8A}{\sin4A}\cdot\frac{\cos2A}{\cos8A}
\displaystyle =\frac{2\sin4A\cos4A}{\sin4A}\cdot\frac{\cos2A}{\cos8A}
\displaystyle =\frac{2\cos4A\cos2A}{\cos8A}
\displaystyle =\frac{\sin8A\cos2A}{\sin4A\cos8A}
\displaystyle =\frac{\tan8A}{\tan2A}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\text{The value of }\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}\text{ is}
\displaystyle \text{(a) }\frac{1}{8}\qquad \text{(b) }\frac{1}{16}\qquad \text{(c) }\frac{1}{32}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2^6\sin\frac{\pi}{65}\prod_{r=0}^{5}\cos\frac{2^r\pi}{65}=\sin\frac{64\pi}{65}
\displaystyle \therefore 64\sin\frac{\pi}{65}\prod_{r=0}^{5}\cos\frac{2^r\pi}{65}=\sin\frac{\pi}{65}
\displaystyle \therefore \prod_{r=0}^{5}\cos\frac{2^r\pi}{65}=\frac{1}{64}
\displaystyle \therefore \text{Correct option is (d) none of these.}
\\

\displaystyle \textbf{Question 4. }\text{If }\cos2x+2\cos x=1\text{ then, }(2-\cos^2x)\sin^2x\text{ is equal to}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }-\sqrt5\qquad \text{(d) }\sqrt5
\displaystyle \text{Answer:}
\displaystyle \cos2x=2\cos^2x-1
\displaystyle 2\cos^2x-1+2\cos x=1
\displaystyle \cos^2x+\cos x-1=0
\displaystyle \therefore \cos^2x=1-\cos x
\displaystyle (2-\cos^2x)\sin^2x=(1+\cos x)(1-\cos^2x)
\displaystyle =(1+\cos x)^2(1-\cos x)
\displaystyle =(1+\cos x)\cos^2x
\displaystyle =\cos^2x+\cos^3x=1
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 5. }\text{For all real values of }x,\ \cot x-2\cot2x\text{ is equal to}
\displaystyle \text{(a) }\tan2x\qquad \text{(b) }\tan x\qquad \text{(c) }-\cot3x\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cot2x=\frac{\cot^2x-1}{2\cot x}
\displaystyle \cot x-2\cot2x=\cot x-\frac{\cot^2x-1}{\cot x}
\displaystyle =\frac{1}{\cot x}=\tan x
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 6. }\text{The value of }2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }\sqrt5\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2\tan\frac{\pi}{10}+3\sec\frac{\pi}{10}-4\cos\frac{\pi}{10}
\displaystyle =\frac{2\sin18^\circ+3-4\cos^218^\circ}{\cos18^\circ}
\displaystyle =\frac{2\sin18^\circ+3-4(1-\sin^218^\circ)}{\cos18^\circ}
\displaystyle =\frac{4\sin^218^\circ+2\sin18^\circ-1}{\cos18^\circ}=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 7. }\text{If in a }\Delta ABC,\ \tan A+\tan B+\tan C=0,\text{ then }\cot A\cot B\cot C=
\displaystyle \text{(a) }6\qquad \text{(b) }1\qquad \text{(c) }\frac{1}{6}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 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=0
\displaystyle \therefore \cot A\cot B\cot C\text{ is not defined}
\displaystyle \therefore \text{Correct option is (d) none of these.}
\\

\displaystyle \textbf{Question 8. }\text{If }\cos\theta=\frac{1}{2}\left(a+\frac{1}{a}\right),\text{ and }\cos3\theta=\lambda\left(a^3+\frac{1}{a^3}\right),\text{ then }\lambda=
\displaystyle \text{(a) }\frac{1}{4}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos3\theta=4\cos^3\theta-3\cos\theta
\displaystyle =4\left(\frac{a+\frac{1}{a}}{2}\right)^3-3\left(\frac{a+\frac{1}{a}}{2}\right)
\displaystyle =\frac{1}{2}\left[\left(a+\frac{1}{a}\right)^3-3\left(a+\frac{1}{a}\right)\right]
\displaystyle =\frac{1}{2}\left(a^3+\frac{1}{a^3}\right)
\displaystyle \therefore \lambda=\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 9. }\text{If }2\tan\alpha=3\tan\beta,\text{ then }\tan(\alpha-\beta)=
\displaystyle \text{(a) }\frac{\sin2\beta}{5-\cos2\beta}\qquad \text{(b) }\frac{\cos2\beta}{5-\cos2\beta}\qquad \text{(c) }\frac{\sin2\beta}{5+\cos2\beta}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2\tan\alpha=3\tan\beta
\displaystyle \therefore \tan\alpha=\frac{3}{2}\tan\beta
\displaystyle \tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
\displaystyle =\frac{\frac{1}{2}\tan\beta}{1+\frac{3}{2}\tan^2\beta}
\displaystyle =\frac{\tan\beta}{2+3\tan^2\beta}
\displaystyle =\frac{\sin\beta\cos\beta}{2\cos^2\beta+3\sin^2\beta}
\displaystyle =\frac{\sin2\beta}{5-\cos2\beta}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 10. }\text{If }\tan\alpha=\frac{1-\cos\beta}{\sin\beta},\text{ then}
\displaystyle \text{(a) }\tan3\alpha=\tan2\beta\qquad \text{(b) }\tan2\alpha=\tan\beta
\displaystyle \text{(c) }\tan2\beta=\tan\alpha\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{1-\cos\beta}{\sin\beta}=\tan\frac{\beta}{2}
\displaystyle \therefore \tan\alpha=\tan\frac{\beta}{2}
\displaystyle \therefore \alpha=\frac{\beta}{2}
\displaystyle \therefore \tan2\alpha=\tan\beta
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 11. }\text{If }\sin\alpha+\sin\beta=a\text{ and }\cos\alpha-\cos\beta=b,\text{ then }\tan\frac{\alpha-\beta}{2}=
\displaystyle \text{(a) }-\frac{a}{b}\qquad \text{(b) }-\frac{b}{a}\qquad \text{(c) }\sqrt{a^2+b^2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin\alpha+\sin\beta=2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=a
\displaystyle \cos\alpha-\cos\beta=-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}=b
\displaystyle \frac{b}{a}=-\tan\frac{\alpha-\beta}{2}
\displaystyle \therefore \tan\frac{\alpha-\beta}{2}=-\frac{b}{a}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 12. }\text{The value of }\left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2(1-2\tan x\cot2x)\text{ is}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }4
\displaystyle \text{Answer:}
\displaystyle \cot\frac{x}{2}-\tan\frac{x}{2}=2\cot x
\displaystyle 1-2\tan x\cot2x=1-2\tan x\cdot\frac{1-\tan^2x}{2\tan x}
\displaystyle =1-(1-\tan^2x)=\tan^2x
\displaystyle \therefore \left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2(1-2\tan x\cot2x)
\displaystyle =(2\cot x)^2\tan^2x=4
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 13. }\text{The value of }\tan\theta\sin\left(\frac{\pi}{2}+\theta\right)\cos\left(\frac{\pi}{2}-\theta\right)\text{ is}
\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }\frac{1}{2}\sin2\theta\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin\left(\frac{\pi}{2}+\theta\right)=\cos\theta
\displaystyle \cos\left(\frac{\pi}{2}-\theta\right)=\sin\theta
\displaystyle \therefore \tan\theta\sin\left(\frac{\pi}{2}+\theta\right)\cos\left(\frac{\pi}{2}-\theta\right)
\displaystyle =\frac{\sin\theta}{\cos\theta}\cdot\cos\theta\cdot\sin\theta=\sin^2\theta
\displaystyle \therefore \text{Correct option is (d) none of these.}
\\

\displaystyle \textbf{Question 14. }\text{The value of }\sin^2\left(\frac{\pi}{18}\right)+\sin^2\left(\frac{\pi}{9}\right)+\sin^2\left(\frac{7\pi}{18}\right)+\sin^2\left(\frac{4\pi}{9}\right)\text{ is}
\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }4\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin^2\frac{\pi}{18}+\sin^2\frac{4\pi}{9}=1
\displaystyle \sin^2\frac{\pi}{9}+\sin^2\frac{7\pi}{18}=1
\displaystyle \therefore \text{Required value}=2
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 15. }\text{If }5\sin\alpha=3\sin(\alpha+2\beta)\neq0,\text{ then }\tan(\alpha+\beta)\text{ is equal to}
\displaystyle \text{(a) }2\tan\beta\qquad \text{(b) }3\tan\beta\qquad \text{(c) }4\tan\beta\qquad \text{(d) }6\tan\beta
\displaystyle \text{Answer:}
\displaystyle 5\sin\alpha=3\sin(\alpha+2\beta)
\displaystyle 5\sin\alpha=3\{\sin\alpha\cos2\beta+\cos\alpha\sin2\beta\}
\displaystyle \sin\alpha(5-3\cos2\beta)=3\cos\alpha\sin2\beta
\displaystyle \tan\alpha=\frac{3\sin2\beta}{5-3\cos2\beta}
\displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
\displaystyle =4\tan\beta
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 16. }\text{The value of }2\cos\theta-\cos3\theta-\cos5\theta-16\cos^3\theta\sin^2\theta\text{ is}
\displaystyle \text{(a) }2\qquad \text{(b) }1\qquad \text{(c) }0\qquad \text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \cos3\theta+\cos5\theta=2\cos4\theta\cos\theta
\displaystyle 2\cos\theta-\cos3\theta-\cos5\theta=2\cos\theta(1-\cos4\theta)
\displaystyle =4\cos\theta\sin^22\theta
\displaystyle =16\cos^3\theta\sin^2\theta
\displaystyle \therefore 2\cos\theta-\cos3\theta-\cos5\theta-16\cos^3\theta\sin^2\theta=0
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 17. }\text{If }A=2\sin^2\theta-\cos2\theta,\text{ then }A\text{ lies in the interval}
\displaystyle \text{(a) }[-1,3]\qquad \text{(b) }[1,2]\qquad \text{(c) }[-2,4]\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle A=2\sin^2\theta-(1-2\sin^2\theta)
\displaystyle =4\sin^2\theta-1
\displaystyle 0\leq\sin^2\theta\leq1
\displaystyle \therefore -1\leq A\leq3
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 18. }\text{The value of }\frac{\cos3\theta}{2\cos2\theta-1}\text{ is equal to}
\displaystyle \text{(a) }\cos\theta\qquad \text{(b) }\sin\theta\qquad \text{(c) }\tan\theta\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2\cos2\theta-1=2(2\cos^2\theta-1)-1
\displaystyle =4\cos^2\theta-3
\displaystyle \cos3\theta=4\cos^3\theta-3\cos\theta
\displaystyle =\cos\theta(4\cos^2\theta-3)
\displaystyle \therefore \frac{\cos3\theta}{2\cos2\theta-1}=\cos\theta
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 19. }\text{If }\tan\left(\frac{\pi}{4}+\theta\right)+\tan\left(\frac{\pi}{4}-\theta\right)=\lambda\sec2\theta,\text{ then}
\displaystyle \text{(a) }3\qquad \text{(b) }4\qquad \text{(c) }1\qquad \text{(d) }2
\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{2(1+\tan^2\theta)}{1-\tan^2\theta}
\displaystyle =2\sec2\theta
\displaystyle \therefore \lambda=2
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 20. }\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) }\frac{1}{2}\cos2\theta\qquad \text{(b) }0\qquad \text{(c) }-\frac{1}{2}\cos2\theta\qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle \cos^2A-\sin^2B=\frac{1+\cos2A}{2}-\frac{1-\cos2B}{2}
\displaystyle =\frac{\cos2A+\cos2B}{2}
\displaystyle =\frac{\cos\left(\frac{\pi}{3}+2\theta\right)+\cos\left(\frac{\pi}{3}-2\theta\right)}{2}
\displaystyle =\frac{2\cos\frac{\pi}{3}\cos2\theta}{2}
\displaystyle =\frac{1}{2}\cos2\theta
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 21. }\frac{\sin3\theta}{1+2\cos2\theta}\text{ is equal to}
\displaystyle \text{(a) }\cos\theta\qquad \text{(b) }\sin\theta\qquad \text{(c) }-\cos\theta\qquad \text{(d) }-\sin\theta
\displaystyle \text{Answer:}
\displaystyle 1+2\cos2\theta=1+2(1-2\sin^2\theta)
\displaystyle =3-4\sin^2\theta
\displaystyle \sin3\theta=3\sin\theta-4\sin^3\theta
\displaystyle =\sin\theta(3-4\sin^2\theta)
\displaystyle \therefore \frac{\sin3\theta}{1+2\cos2\theta}=\sin\theta
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 22. }\text{The value of }2\sin^2B+4\cos(A+B)\sin A\sin B+\cos2(A+B)\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }\cos3A\qquad \text{(c) }\cos2A\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2\sin^2B=1-\cos2B
\displaystyle \cos2(A+B)-\cos2B=-2\sin(A+2B)\sin A
\displaystyle \therefore \text{Expression}=1-2\sin A\sin(A+2B)+4\cos(A+B)\sin A\sin B
\displaystyle =1+2\sin A\{-\sin(A+2B)+2\cos(A+B)\sin B\}
\displaystyle =1+2\sin A\{-\sin(A+2B)+\sin(A+2B)-\sin A\}
\displaystyle =1-2\sin^2A=\cos2A
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 23. }\text{The value of }\frac{2(\sin2\theta+2\cos^2\theta-1)}{\cos\theta-\sin\theta-\cos3\theta+\sin3\theta}\text{ is}
\displaystyle \text{(a) }\cos\theta\qquad \text{(b) }\sec\theta\qquad \text{(c) }\mathrm{cosec}\ \theta\qquad \text{(d) }\sin\theta
\displaystyle \text{Answer:}
\displaystyle \sin2\theta+2\cos^2\theta-1=\sin2\theta+\cos2\theta
\displaystyle \text{Numerator}=2(\sin2\theta+\cos2\theta)
\displaystyle \cos\theta-\cos3\theta=2\sin2\theta\sin\theta
\displaystyle \sin3\theta-\sin\theta=2\cos2\theta\sin\theta
\displaystyle \text{Denominator}=2\sin\theta(\sin2\theta+\cos2\theta)
\displaystyle \therefore \text{Required value}=\frac{1}{\sin\theta}=\mathrm{cosec}\ \theta
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 24. }2(1-2\sin^27\theta)\sin3\theta\text{ is equal to}
\displaystyle \text{(a) }\sin17\theta-\sin11\theta\qquad \text{(b) }\sin11\theta-\sin17\theta
\displaystyle \text{(c) }\cos17\theta-\cos11\theta\qquad \text{(d) }\cos17\theta+\cos11\theta
\displaystyle \text{Answer:}
\displaystyle 1-2\sin^27\theta=\cos14\theta
\displaystyle 2(1-2\sin^27\theta)\sin3\theta=2\cos14\theta\sin3\theta
\displaystyle =\sin17\theta-\sin11\theta
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 25. }\text{If }\alpha\text{ and }\beta\text{ are acute angles satisfying }\cos2\alpha=\frac{3\cos2\beta-1}{3-\cos2\beta},\text{ then }\tan\alpha=
\displaystyle \text{(a) }\sqrt2\tan\beta\qquad \text{(b) }\frac{1}{\sqrt2}\tan\beta\qquad \text{(c) }\sqrt2\cot\beta\qquad \text{(d) }\frac{1}{\sqrt2}\cot\beta
\displaystyle \text{Answer:}
\displaystyle \frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{3\cos2\beta-1}{3-\cos2\beta}
\displaystyle \text{Let }t=\tan^2\beta,\text{ then }\cos2\beta=\frac{1-t}{1+t}
\displaystyle \frac{3\cos2\beta-1}{3-\cos2\beta}=\frac{1-2t}{1+2t}
\displaystyle \therefore \frac{1-\tan^2\alpha}{1+\tan^2\alpha}=\frac{1-2\tan^2\beta}{1+2\tan^2\beta}
\displaystyle \therefore \tan^2\alpha=2\tan^2\beta
\displaystyle \therefore \tan\alpha=\sqrt2\tan\beta
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 26. }\text{If }\tan\frac{\theta}{2}=\sqrt{\frac{1-e}{1+e}}\tan\frac{\alpha}{2},\text{ then }\cos\alpha=
\displaystyle \text{(a) }1-e\cos(\cos\theta+e)\qquad \text{(b) }\frac{1+e\cos\theta}{\cos\theta-e}
\displaystyle \text{(c) }\frac{1-e\cos\theta}{\cos\theta-e}\qquad \text{(d) }\frac{\cos\theta-e}{1-e\cos\theta}
\displaystyle \text{Answer:}
\displaystyle \tan^2\frac{\theta}{2}=\frac{1-e}{1+e}\tan^2\frac{\alpha}{2}
\displaystyle \frac{1-\cos\theta}{1+\cos\theta}=\frac{1-e}{1+e}\cdot\frac{1-\cos\alpha}{1+\cos\alpha}
\displaystyle \frac{1-\cos\alpha}{1+\cos\alpha}=\frac{1+e}{1-e}\cdot\frac{1-\cos\theta}{1+\cos\theta}
\displaystyle \therefore \cos\alpha=\frac{\cos\theta-e}{1-e\cos\theta}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 27. }\text{If }(2^n+1)\theta=\pi,\text{ then }2^n\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=
\displaystyle \text{(a) }-1\qquad \text{(b) }1\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2^n\sin\theta\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=\sin2^n\theta
\displaystyle 2^n\cos\theta\cos2\theta\cos2^2\theta\cdots\cos2^{n-1}\theta=\frac{\sin2^n\theta}{\sin\theta}
\displaystyle (2^n+1)\theta=\pi\Rightarrow 2^n\theta=\pi-\theta
\displaystyle \therefore \frac{\sin2^n\theta}{\sin\theta}=\frac{\sin(\pi-\theta)}{\sin\theta}=1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 28. }\text{If }\tan\theta=t,\text{ then }\tan2\theta+\sec2\theta\text{ is equal to}
\displaystyle \text{(a) }\frac{1+t}{1-t}\qquad \text{(b) }\frac{1-t}{1+t}\qquad \text{(c) }\frac{2t}{1-t}\qquad \text{(d) }\frac{2t}{1+t}
\displaystyle \text{Answer:}
\displaystyle \tan2\theta+\sec2\theta=\frac{\sin2\theta+1}{\cos2\theta}
\displaystyle =\frac{\frac{2t}{1+t^2}+1}{\frac{1-t^2}{1+t^2}}
\displaystyle =\frac{(1+t)^2}{(1-t)(1+t)}
\displaystyle =\frac{1+t}{1-t}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 29. }\text{The value of }\cos^4\theta+\sin^4\theta-6\cos^2\theta\sin^2\theta\text{ is}
\displaystyle \text{(a) }\cos2\theta\qquad \text{(b) }\sin2\theta\qquad \text{(c) }\cos4\theta\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos^4\theta+\sin^4\theta-6\cos^2\theta\sin^2\theta
\displaystyle =(\cos^2\theta+\sin^2\theta)^2-8\cos^2\theta\sin^2\theta
\displaystyle =1-2\sin^22\theta
\displaystyle =\cos4\theta
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 30. }\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) }\cos2A\qquad \text{(b) }\sin2A\qquad \text{(c) }\cos A\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \cos(36^\circ-A)\cos(36^\circ+A)=\frac{\cos72^\circ+\cos2A}{2}
\displaystyle \cos(54^\circ-A)\cos(54^\circ+A)=\frac{\cos108^\circ+\cos2A}{2}
\displaystyle \therefore \text{Required value}=\frac{\cos72^\circ+\cos108^\circ+2\cos2A}{2}
\displaystyle \cos108^\circ=-\cos72^\circ
\displaystyle \therefore \text{Required value}=\cos2A
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 31. }\text{The value of }\tan\theta\tan(60^\circ-\theta)\tan(60^\circ+\theta)\text{ is}
\displaystyle \text{(a) }\cot3\theta\qquad \text{(b) }2\cot3\theta\qquad \text{(c) }\tan3\theta\qquad \text{(d) }3\tan3\theta
\displaystyle \text{Answer:}
\displaystyle \tan3\theta=\frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}
\displaystyle \tan\theta\tan(60^\circ-\theta)\tan(60^\circ+\theta)=\tan3\theta
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 32. }\text{The value of }\tan\theta+\tan(60^\circ+\theta)+\tan(120^\circ+\theta)\text{ is}
\displaystyle \text{(a) }3\tan3\theta\qquad \text{(b) }\tan3\theta\qquad \text{(c) }3\cot3\theta\qquad \text{(d) }\cot3\theta
\displaystyle \text{Answer:}
\displaystyle \tan A+\tan B+\tan C=\tan A\tan B\tan C,\text{ if }A+B+C=180^\circ
\displaystyle \theta+(60^\circ+\theta)+(120^\circ+\theta)=180^\circ+3\theta
\displaystyle \tan\theta+\tan(60^\circ+\theta)+\tan(120^\circ+\theta)=3\tan3\theta
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 33. }\text{The value of }\frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}\text{ is}
\displaystyle \text{(a) }\cot\frac{\alpha}{2}\qquad \text{(b) }\cot\alpha\qquad \text{(c) }\tan\frac{\alpha}{2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin5\alpha-\sin3\alpha=2\cos4\alpha\sin\alpha
\displaystyle \cos5\alpha+\cos3\alpha=2\cos4\alpha\cos\alpha
\displaystyle \therefore \cos5\alpha+2\cos4\alpha+\cos3\alpha=2\cos4\alpha(1+\cos\alpha)
\displaystyle \therefore \frac{\sin5\alpha-\sin3\alpha}{\cos5\alpha+2\cos4\alpha+\cos3\alpha}=\frac{\sin\alpha}{1+\cos\alpha}
\displaystyle =\tan\frac{\alpha}{2}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 34. }\frac{\sin5\theta}{\sin\theta}\text{ is equal to}
\displaystyle \text{(a) }16\cos^4\theta-12\cos^2\theta+1\qquad \text{(b) }16\cos^4\theta+12\cos^2\theta+1
\displaystyle \text{(c) }16\cos^4\theta-12\cos^2\theta-1\qquad \text{(d) }16\cos^4\theta+12\cos^2\theta-1
\displaystyle \text{Answer:}
\displaystyle \sin5\theta=16\sin^5\theta-20\sin^3\theta+5\sin\theta
\displaystyle \frac{\sin5\theta}{\sin\theta}=16\sin^4\theta-20\sin^2\theta+5
\displaystyle =16(1-\cos^2\theta)^2-20(1-\cos^2\theta)+5
\displaystyle =16\cos^4\theta-12\cos^2\theta+1
\displaystyle \therefore \text{Correct option is (a).}
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\displaystyle \textbf{Question 35. }\text{If }n=1,2,3,\ldots,\text{ then }\cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{n-1}\alpha\text{ is equal to}
\displaystyle \text{(a) }\frac{\sin2^n\alpha}{2n\sin\alpha}\qquad \text{(b) }\frac{\sin2^n\alpha}{2^n\sin2^{n-1}\alpha}
\displaystyle \text{(c) }\frac{\sin4^n-1\alpha}{4^n-1\sin\alpha}\qquad \text{(d) }\frac{\sin2^n\alpha}{2^n\sin\alpha}
\displaystyle \text{Answer:}
\displaystyle 2^n\sin\alpha\cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{n-1}\alpha=\sin2^n\alpha
\displaystyle \therefore \cos\alpha\cos2\alpha\cos4\alpha\cdots\cos2^{n-1}\alpha=\frac{\sin2^n\alpha}{2^n\sin\alpha}
\displaystyle \therefore \text{Correct option is (d).}
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\displaystyle \textbf{Question 36. }\text{If }\tan\theta=\frac{a}{b},\text{ then }b\cos2\theta+a\sin2\theta\text{ is equal to}
\displaystyle \text{(a) }a\qquad \text{(b) }b\qquad \text{(c) }\frac{a}{b}\qquad \text{(d) }\frac{b}{a}
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac{a}{b}
\displaystyle \cos2\theta=\frac{b^2-a^2}{a^2+b^2},\quad \sin2\theta=\frac{2ab}{a^2+b^2}
\displaystyle b\cos2\theta+a\sin2\theta=b\cdot\frac{b^2-a^2}{a^2+b^2}+a\cdot\frac{2ab}{a^2+b^2}
\displaystyle =\frac{b^3-a^2b+2a^2b}{a^2+b^2}
\displaystyle =\frac{b(a^2+b^2)}{a^2+b^2}=b
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 37. }\text{If }\tan\alpha=\frac{1}{7},\ \tan\beta=\frac{1}{3},\text{ then }\cos2\alpha\text{ is equal to}
\displaystyle \text{(a) }\sin2\beta\qquad \text{(b) }\sin4\beta\qquad \text{(c) }\sin3\beta\qquad \text{(d) }\cos2\beta
\displaystyle \text{Answer:}
\displaystyle \cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}
\displaystyle =\frac{1-\frac{1}{49}}{1+\frac{1}{49}}=\frac{48}{50}=\frac{24}{25}
\displaystyle \sin2\beta=\frac{2\tan\beta}{1+\tan^2\beta}
\displaystyle =\frac{2\cdot\frac{1}{3}}{1+\frac{1}{9}}=\frac{3}{5}
\displaystyle \cos2\beta=\frac{1-\frac{1}{9}}{1+\frac{1}{9}}=\frac{4}{5}
\displaystyle \sin4\beta=2\sin2\beta\cos2\beta=2\cdot\frac{3}{5}\cdot\frac{4}{5}=\frac{24}{25}
\displaystyle \therefore \cos2\alpha=\sin4\beta
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 38. }\text{The value of }\cos^248^\circ-\sin^212^\circ\text{ is}
\displaystyle \text{(a) }\frac{\sqrt5+1}{8}\qquad \text{(b) }\frac{\sqrt5-1}{8}\qquad \text{(c) }\frac{\sqrt5+1}{5}\qquad \text{(d) }\frac{\sqrt5+1}{2\sqrt2}
\displaystyle \text{Answer:}
\displaystyle \cos^248^\circ-\sin^212^\circ
\displaystyle =\frac{1+\cos96^\circ}{2}-\frac{1-\cos24^\circ}{2}
\displaystyle =\frac{\cos96^\circ+\cos24^\circ}{2}
\displaystyle =\frac{2\cos60^\circ\cos36^\circ}{2}
\displaystyle =\frac{\cos36^\circ}{2}
\displaystyle =\frac{1}{2}\cdot\frac{\sqrt5+1}{4}=\frac{\sqrt5+1}{8}
\displaystyle \therefore \text{Correct option is (a).}
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