\displaystyle \text{Transformation Formulae}

\displaystyle \text{Values of Trigonometric Functions at Multiples and Submultiples of a Number}

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

\displaystyle \textbf{Question 1: }\cos40^\circ+\cos80^\circ+\cos160^\circ+\cos240^\circ=
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }\frac12\qquad\text{(d) }-\frac12
\displaystyle \text{Answer:}
\displaystyle \cos40^\circ+\cos80^\circ
\displaystyle =2\cos60^\circ\cos20^\circ=\cos20^\circ.
\displaystyle \cos160^\circ=\cos(180^\circ-20^\circ)=-\cos20^\circ.
\displaystyle \therefore \cos40^\circ+\cos80^\circ+\cos160^\circ=0.
\displaystyle \cos240^\circ=\cos(180^\circ+60^\circ)=-\frac12.
\displaystyle \therefore \text{Required value}=-\frac12.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ=
\displaystyle \text{(a) }0\qquad\text{(b) }\frac12\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin163^\circ=\sin17^\circ,\qquad\cos347^\circ=\cos13^\circ.
\displaystyle \sin73^\circ=\cos17^\circ,\qquad\sin167^\circ=\sin13^\circ.
\displaystyle \therefore \text{Required expression}
\displaystyle =\sin17^\circ\cos13^\circ+\cos17^\circ\sin13^\circ
\displaystyle =\sin(17^\circ+13^\circ)
\displaystyle =\sin30^\circ=\frac12.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\sin2\theta+\sin2\phi=\frac12\text{ and }\cos2\theta+\cos2\phi=\frac32,\text{ then}
\displaystyle \cos^2(\theta-\phi)=
\displaystyle \text{(a) }\frac38\qquad\text{(b) }\frac58\qquad\text{(c) }\frac34\qquad\text{(d) }\frac54
\displaystyle \text{Answer:}
\displaystyle \sin2\theta+\sin2\phi
\displaystyle =2\sin(\theta+\phi)\cos(\theta-\phi)=\frac12.
\displaystyle \cos2\theta+\cos2\phi
\displaystyle =2\cos(\theta+\phi)\cos(\theta-\phi)=\frac32.
\displaystyle \text{Squaring and adding,}
\displaystyle 4\cos^2(\theta-\phi)\{\sin^2(\theta+\phi)+\cos^2(\theta+\phi)\}
\displaystyle =\frac14+\frac94=\frac52.
\displaystyle \therefore 4\cos^2(\theta-\phi)=\frac52.
\displaystyle \therefore \cos^2(\theta-\phi)=\frac58.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The value of }\cos52^\circ+\cos68^\circ+\cos172^\circ\text{ is}
\displaystyle \text{(a) }0\qquad\text{(b) }1\qquad\text{(c) }2\qquad\text{(d) }\frac32
\displaystyle \text{Answer:}
\displaystyle \cos52^\circ+\cos68^\circ
\displaystyle =2\cos60^\circ\cos(-8^\circ)
\displaystyle =\cos8^\circ.
\displaystyle \cos172^\circ=\cos(180^\circ-8^\circ)=-\cos8^\circ.
\displaystyle \therefore \cos52^\circ+\cos68^\circ+\cos172^\circ=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The value of }\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ\text{ is}
\displaystyle \text{(a) }\frac12\qquad\text{(b) }-\frac12\qquad\text{(c) }-1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ
\displaystyle =(\sin78^\circ-\sin42^\circ)-(\sin66^\circ-\sin6^\circ).
\displaystyle \sin78^\circ-\sin42^\circ
\displaystyle =2\cos60^\circ\sin18^\circ=\sin18^\circ.
\displaystyle \sin66^\circ-\sin6^\circ
\displaystyle =2\cos36^\circ\sin30^\circ=\cos36^\circ.
\displaystyle \therefore \text{Required expression}=\sin18^\circ-\cos36^\circ.
\displaystyle \sin18^\circ=\frac{\sqrt5-1}{4},\qquad\cos36^\circ=\frac{\sqrt5+1}{4}.
\displaystyle \therefore \text{Required expression}
\displaystyle =\frac{\sqrt5-1}{4}-\frac{\sqrt5+1}{4}=-\frac12.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }\sin\alpha+\sin\beta=a\text{ and }\cos\alpha-\cos\beta=b,\text{ then }  \tan\left(\frac{\alpha-\beta}{2}\right)=
\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
\displaystyle =2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)=a.
\displaystyle \cos\alpha-\cos\beta
\displaystyle =-2\sin\left(\frac{\alpha+\beta}{2}\right)\sin\left(\frac{\alpha-\beta}{2}\right)=b.
\displaystyle \text{Dividing the second equation by the first,}
\displaystyle \frac{b}{a}=-\tan\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \therefore \tan\left(\frac{\alpha-\beta}{2}\right)=-\frac{b}{a}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\cos35^\circ+\cos85^\circ+\cos155^\circ=
\displaystyle \text{(a) }0\qquad\text{(b) }\frac{1}{\sqrt3}\qquad\text{(c) }\frac{1}{\sqrt2}\qquad\text{(d) }\cos275^\circ
\displaystyle \text{Answer:}
\displaystyle \cos35^\circ+\cos155^\circ
\displaystyle =2\cos\left(\frac{35^\circ+155^\circ}{2}\right)
\displaystyle \quad\cos\left(\frac{35^\circ-155^\circ}{2}\right)
\displaystyle =2\cos95^\circ\cos(-60^\circ)=\cos95^\circ.
\displaystyle \cos95^\circ=-\cos85^\circ.
\displaystyle \therefore \cos35^\circ+\cos85^\circ+\cos155^\circ=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The value of }\sin50^\circ-\sin70^\circ+\sin10^\circ\text{ is equal to}
\displaystyle \text{(a) }1\qquad\text{(b) }0\qquad\text{(c) }\frac12\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \sin50^\circ+\sin10^\circ
\displaystyle =2\sin\left(\frac{50^\circ+10^\circ}{2}\right)
\displaystyle \quad\cos\left(\frac{50^\circ-10^\circ}{2}\right)
\displaystyle =2\sin30^\circ\cos20^\circ=\cos20^\circ.
\displaystyle \cos20^\circ=\sin70^\circ.
\displaystyle \therefore \sin50^\circ-\sin70^\circ+\sin10^\circ=0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ\text{ is equal to}
\displaystyle \text{(a) }\sin36^\circ\qquad\text{(b) }\cos36^\circ\qquad\text{(c) }\sin7^\circ\qquad\text{(d) }\cos7^\circ
\displaystyle \text{Answer:}
\displaystyle \sin47^\circ-\sin11^\circ=2\cos29^\circ\sin18^\circ.
\displaystyle \sin61^\circ-\sin25^\circ=2\cos43^\circ\sin18^\circ.
\displaystyle \therefore \text{Required expression}
\displaystyle =2\sin18^\circ(\cos29^\circ+\cos43^\circ)
\displaystyle =4\sin18^\circ\cos36^\circ\cos7^\circ.
\displaystyle 4\sin18^\circ\cos36^\circ=1.
\displaystyle \therefore \text{Required expression}=\cos7^\circ.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\cos A=m\cos B,\text{ then } \cot\left(\frac{A+B}{2}\right)\cot\left(\frac{B-A}{2}\right)=
\displaystyle \text{(a) }\frac{m-1}{m+1}\qquad\text{(b) }\frac{m+2}{m-2}\qquad\text{(c) }\frac{m+1}{m-1}
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos A=m\cos B.
\displaystyle \cos A+\cos B=(m+1)\cos B.
\displaystyle \cos A-\cos B=(m-1)\cos B.
\displaystyle \therefore \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}.
\displaystyle \cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right).
\displaystyle \cos A-\cos B=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right).
\displaystyle \therefore \frac{\cos A+\cos B}{\cos A-\cos B}
\displaystyle =-\cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right).
\displaystyle \text{But }\cot\left(\frac{B-A}{2}\right)=-\cot\left(\frac{A-B}{2}\right).
\displaystyle \therefore \cot\left(\frac{A+B}{2}\right)\cot\left(\frac{B-A}{2}\right)
\displaystyle =\frac{m+1}{m-1}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }A,B,C\text{ are in A.P., then }\frac{\sin A-\sin C}{\cos C-\cos A}=
\displaystyle \text{(a) }\tan B\qquad\text{(b) }\cot B\qquad\text{(c) }\tan2B\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle A,B,C\text{ are in A.P.}\Rightarrow A+C=2B.
\displaystyle \sin A-\sin C
\displaystyle =2\cos\left(\frac{A+C}{2}\right)\sin\left(\frac{A-C}{2}\right)
\displaystyle =2\cos B\sin\left(\frac{A-C}{2}\right).
\displaystyle \cos C-\cos A
\displaystyle =2\sin\left(\frac{A+C}{2}\right)\sin\left(\frac{A-C}{2}\right)
\displaystyle =2\sin B\sin\left(\frac{A-C}{2}\right).
\displaystyle \therefore \frac{\sin A-\sin C}{\cos C-\cos A}=\frac{\cos B}{\sin B}=\cot B.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\sin(B+C-A),\ \sin(C+A-B),\ \sin(A+B-C)
\displaystyle \text{are in A.P., then }\cot A,\cot B,\cot C\text{ are in}
\displaystyle \text{(a) G.P.}\qquad\text{(b) H.P.}\qquad\text{(c) A.P.}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Since the given terms are in A.P.,}
\displaystyle 2\sin(C+A-B)=\sin(B+C-A)+\sin(A+B-C).
\displaystyle \sin(B+C-A)+\sin(A+B-C)
\displaystyle =2\sin B\cos(C-A).
\displaystyle \therefore \sin(A+C-B)=\sin B\cos(C-A).
\displaystyle \sin(A+C)\cos B-\cos(A+C)\sin B
\displaystyle =\sin B\cos(C-A).
\displaystyle \therefore \sin(A+C)\cos B
\displaystyle =\sin B\{\cos(A+C)+\cos(C-A)\}.
\displaystyle \cos(A+C)+\cos(C-A)=2\cos A\cos C.
\displaystyle \therefore \sin(A+C)\cos B=2\sin B\cos A\cos C.
\displaystyle (\sin A\cos C+\cos A\sin C)\cos B
\displaystyle =2\sin B\cos A\cos C.
\displaystyle \text{Dividing by }\cos A\cos B\cos C,
\displaystyle \tan A+\tan C=2\tan B.
\displaystyle \therefore \tan A,\tan B,\tan C\text{ are in A.P.}
\displaystyle \therefore \cot A,\cot B,\cot C\text{ are in H.P.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\sin x+\sin y=\sqrt3(\cos y-\cos x),\text{ then }\sin3x+\sin3y=
\displaystyle \text{(a) }2\sin3x\qquad\text{(b) }0\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin x+\sin y=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right).
\displaystyle \cos y-\cos x=2\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right).
\displaystyle \therefore 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)
\displaystyle =2\sqrt3\sin\left(\frac{x+y}{2}\right)\sin\left(\frac{x-y}{2}\right).
\displaystyle \therefore \cot\left(\frac{x-y}{2}\right)=\sqrt3.
\displaystyle \Rightarrow \frac{x-y}{2}=\frac{\pi}{6}\pmod{\pi}.
\displaystyle \therefore 3(x-y)=\pi\pmod{6\pi}.
\displaystyle \sin3x+\sin3y=2\sin\frac{3(x+y)}{2}\cos\frac{3(x-y)}{2}.
\displaystyle \text{But }\cos\frac{3(x-y)}{2}=\cos\frac{\pi}{2}=0.
\displaystyle \therefore \sin3x+\sin3y=0.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\tan\alpha=\frac{x}{x+1}\text{ and }\tan\beta=\frac{1}{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}+\frac{1}{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 \textbf{Question 15: }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 8\sin\frac{x}{8}\cos\frac{x}{8}\cos\frac{x}{4}\cos\frac{x}{2}
\displaystyle =4\sin\frac{x}{4}\cos\frac{x}{4}\cos\frac{x}{2}
\displaystyle =2\sin\frac{x}{2}\cos\frac{x}{2}
\displaystyle =\sin x.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\frac{\sec8A-1}{\sec4A-1}\text{ is equal to}
\displaystyle \text{(a) }\frac{\tan2A}{\tan8A}\qquad\text{(b) }\frac{\tan8A}{\tan2A}
\displaystyle \text{(c) }\frac{\cot8A}{\cot2A}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \frac{\sec8A-1}{\sec4A-1}
\displaystyle =\frac{(1-\cos8A)\cos4A}{(1-\cos4A)\cos8A}.
\displaystyle =\frac{2\sin^24A\cos4A}{2\sin^22A\cos8A}
\displaystyle =\frac{4\cos^22A\cos4A}{\cos8A}.
\displaystyle \text{Also, }\frac{\tan8A}{\tan2A}
\displaystyle =\frac{\sin8A\cos2A}{\cos8A\sin2A}.
\displaystyle \sin8A=2\sin4A\cos4A
\displaystyle =4\sin2A\cos2A\cos4A.
\displaystyle \therefore \frac{\tan8A}{\tan2A}
\displaystyle =\frac{4\cos^22A\cos4A}{\cos8A}.
\displaystyle \therefore \frac{\sec8A-1}{\sec4A-1}=\frac{\tan8A}{\tan2A}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\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) }\frac18\qquad\text{(b) }\frac1{16}\qquad\text{(c) }\frac1{32}\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin64x=64\sin x\cos x\cos2x\cos4x\cos8x\cos16x\cos32x.
\displaystyle \text{Putting }x=\frac{\pi}{65},
\displaystyle \sin\frac{64\pi}{65}=64\sin\frac{\pi}{65}\cos\frac{\pi}{65}\cos\frac{2\pi}{65}  \quad\cos\frac{4\pi}{65}\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}.
\displaystyle \sin\frac{64\pi}{65}=\sin\left(\pi-\frac{\pi}{65}\right)=\sin\frac{\pi}{65}.
\displaystyle \therefore 64\cos\frac{\pi}{65}\cos\frac{2\pi}{65}\cos\frac{4\pi}{65}
\displaystyle \quad\cos\frac{8\pi}{65}\cos\frac{16\pi}{65}\cos\frac{32\pi}{65}=1.
\displaystyle \therefore \text{Required value}=\frac1{64}.
\displaystyle \therefore \text{Option (d), none of these, is correct.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\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 x=1.
\displaystyle 2\cos^2x-1+2\cos x=1.
\displaystyle \therefore \cos^2x+\cos x=1.
\displaystyle \therefore 1-\cos^2x=\cos x.
\displaystyle \text{Also, }2-\cos^2x=1+\cos x.
\displaystyle (2-\cos^2x)\sin^2x
\displaystyle =(1+\cos x)(1-\cos^2x)
\displaystyle =\cos x(1+\cos x)
\displaystyle =\cos x+\cos^2x=1.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\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 2\cot2x=2\frac{\cos2x}{\sin2x}
\displaystyle =\frac{\cos^2x-\sin^2x}{\sin x\cos x}
\displaystyle =\cot x-\tan x.
\displaystyle \therefore \cot x-2\cot2x
\displaystyle =\cot x-(\cot x-\tan x)=\tan x.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\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 \text{Let }x=\frac{\pi}{10}.
\displaystyle 3x=\frac{3\pi}{10}=\frac{\pi}{2}-2x.
\displaystyle \therefore \cos3x=\sin2x.
\displaystyle 4\cos^3x-3\cos x=2\sin x\cos x.
\displaystyle \text{Dividing by }\cos x,
\displaystyle 4\cos^2x-3=2\sin x.
\displaystyle \text{Dividing again by }\cos x,
\displaystyle 4\cos x-3\sec x=2\tan x.
\displaystyle \therefore 2\tan x+3\sec x-4\cos x
\displaystyle =2\tan x-(4\cos x-3\sec x)
\displaystyle =2\tan x-2\tan x=0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If in a }\triangle ABC,\ \tan A+\tan B+\tan C=0,\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{In }\triangle ABC,\quad A+B+C=\pi.
\displaystyle \therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C.
\displaystyle \text{Given }\tan A+\tan B+\tan C=0.
\displaystyle \therefore \tan A\tan B\tan C=0.
\displaystyle \text{Thus at least one of }\tan A,\tan B,\tan C\text{ is zero.}
\displaystyle \text{But no angle of a non-degenerate triangle can be }0.
\displaystyle \therefore \text{the given condition is not possible for a triangle.}
\displaystyle \therefore \text{Option (d), none of these, is correct.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }\cos\theta=\frac12\left(a+\frac1a\right),\text{ and}
\displaystyle \cos3\theta=\lambda\left(a^3+\frac1{a^3}\right),\text{ then }\lambda=
\displaystyle \text{(a) }\frac14\qquad\text{(b) }\frac12\qquad\text{(c) }1\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 2\cos\theta=a+\frac1a.
\displaystyle \left(a+\frac1a\right)^3=a^3+\frac1{a^3}+3\left(a+\frac1a\right).
\displaystyle \therefore a^3+\frac1{a^3}=8\cos^3\theta-6\cos\theta.
\displaystyle =2(4\cos^3\theta-3\cos\theta)=2\cos3\theta.
\displaystyle \therefore \cos3\theta=\frac12\left(a^3+\frac1{a^3}\right).
\displaystyle \therefore \lambda=\frac12.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\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}
\displaystyle \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 \Rightarrow \tan\alpha=\frac32\tan\beta.
\displaystyle \tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}
\displaystyle =\frac{\frac32\tan\beta-\tan\beta}{1+\frac32\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{\frac12\sin2\beta}{\frac12(5-\cos2\beta)}
\displaystyle =\frac{\sin2\beta}{5-\cos2\beta}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\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 \tan\alpha=\frac{1-\cos\beta}{\sin\beta}
\displaystyle =\frac{2\sin^2\frac{\beta}{2}}{2\sin\frac{\beta}{2}\cos\frac{\beta}{2}}
\displaystyle =\tan\frac{\beta}{2}.
\displaystyle \therefore \alpha=\frac{\beta}{2}+n\pi,\qquad n\in Z.
\displaystyle \Rightarrow 2\alpha=\beta+2n\pi.
\displaystyle \therefore \tan2\alpha=\tan\beta.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{If }\sin\alpha+\sin\beta=a\text{ and }\cos\alpha-\cos\beta=b,\text{ then}
\displaystyle \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 a=\sin\alpha+\sin\beta
\displaystyle =2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}.
\displaystyle b=\cos\alpha-\cos\beta
\displaystyle =-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}.
\displaystyle \therefore \frac{b}{a}=-\tan\frac{\alpha-\beta}{2}.
\displaystyle \therefore \tan\frac{\alpha-\beta}{2}=-\frac{b}{a}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{The value of }\left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2
\displaystyle \left(1-2\tan x\cot2x\right)\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}
\displaystyle =\frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\sin\frac{x}{2}\cos\frac{x}{2}}
\displaystyle =\frac{\cos x}{\frac12\sin x}=2\cot x.
\displaystyle \therefore \left(\cot\frac{x}{2}-\tan\frac{x}{2}\right)^2=4\cot^2x.
\displaystyle \cot2x=\frac{1-\tan^2x}{2\tan x}.
\displaystyle \therefore 1-2\tan x\cot2x
\displaystyle =1-(1-\tan^2x)=\tan^2x.
\displaystyle \therefore \left(2\cot x\right)^2\tan^2x=4.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{The value of }\tan\theta\sin\left(\frac{\pi}{2}+\theta\right)
\displaystyle \cos\left(\frac{\pi}{2}-\theta\right)\text{ is}
\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad\text{(c) }\frac12\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)
\displaystyle \quad\cos\left(\frac{\pi}{2}-\theta\right)
\displaystyle =\tan\theta\cos\theta\sin\theta
\displaystyle =\frac{\sin\theta}{\cos\theta}\cos\theta\sin\theta
\displaystyle =\sin^2\theta.
\displaystyle \therefore \text{Option (d), none of these, is correct.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{The value of }\sin^2\left(\frac{\pi}{18}\right)+\sin^2\left(\frac{\pi}{9}\right)
\displaystyle +\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 \frac{4\pi}{9}=\frac{\pi}{2}-\frac{\pi}{18}
\displaystyle \Rightarrow \sin^2\frac{4\pi}{9}=\cos^2\frac{\pi}{18}.
\displaystyle \frac{7\pi}{18}=\frac{\pi}{2}-\frac{\pi}{9}
\displaystyle \Rightarrow \sin^2\frac{7\pi}{18}=\cos^2\frac{\pi}{9}.
\displaystyle \therefore \text{Required expression}
\displaystyle =\sin^2\frac{\pi}{18}+\cos^2\frac{\pi}{18}
\displaystyle \quad+\sin^2\frac{\pi}{9}+\cos^2\frac{\pi}{9}
\displaystyle =1+1=2.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{If }5\sin\alpha=3\sin(\alpha+2\beta)\ne0,\text{ then}
\displaystyle \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 \therefore \tan\alpha=\frac{3\sin2\beta}{5-3\cos2\beta}.
\displaystyle =\frac{6\tan\beta}{2+8\tan^2\beta}
\displaystyle =\frac{3\tan\beta}{1+4\tan^2\beta}.
\displaystyle \therefore \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}
\displaystyle =\frac{\frac{3t}{1+4t^2}+t}{1-\frac{3t^2}{1+4t^2}},\qquad t=\tan\beta.
\displaystyle =\frac{4t(1+t^2)}{1+t^2}=4t=4\tan\beta.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{The value of }2\cos\theta-\cos3\theta-\cos5\theta
\displaystyle -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 \therefore 2\cos\theta-\cos3\theta-\cos5\theta
\displaystyle =2\cos\theta(1-\cos4\theta).
\displaystyle =2\cos\theta(2\sin^22\theta)
\displaystyle =4\cos\theta(4\sin^2\theta\cos^2\theta)
\displaystyle =16\cos^3\theta\sin^2\theta.
\displaystyle \therefore 2\cos\theta-\cos3\theta-\cos5\theta
\displaystyle \quad-16\cos^3\theta\sin^2\theta=0.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\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-\cos2\theta.
\displaystyle \text{Using }\cos2\theta=1-2\sin^2\theta,
\displaystyle A=2\sin^2\theta-(1-2\sin^2\theta)
\displaystyle =4\sin^2\theta-1.
\displaystyle \text{Since }0\leq\sin^2\theta\leq1,
\displaystyle 0\leq4\sin^2\theta\leq4.
\displaystyle \therefore -1\leq4\sin^2\theta-1\leq3.
\displaystyle \therefore -1\leq A\leq3.
\displaystyle \therefore A\in[-1,3].
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\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 \cos3\theta=4\cos^3\theta-3\cos\theta.
\displaystyle =\cos\theta(4\cos^2\theta-3).
\displaystyle 2\cos2\theta-1
\displaystyle =2(2\cos^2\theta-1)-1
\displaystyle =4\cos^2\theta-3.
\displaystyle \therefore \cos3\theta=\cos\theta(2\cos2\theta-1).
\displaystyle \therefore \frac{\cos3\theta}{2\cos2\theta-1}=\cos\theta.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{If }\tan\left(\frac{\pi}{4}+\theta\right)+\tan\left(\frac{\pi}{4}-\theta\right)
\displaystyle =\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)+\tan\left(\frac{\pi}{4}-\theta\right)
\displaystyle =\frac{\sin\left(\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{4}+\theta\right)\cos\left(\frac{\pi}{4}-\theta\right)}.
\displaystyle \cos\left(\frac{\pi}{4}+\theta\right)\cos\left(\frac{\pi}{4}-\theta\right)
\displaystyle =\frac12\left(\cos2\theta+\cos\frac{\pi}{2}\right)=\frac12\cos2\theta.
\displaystyle \therefore \text{Required expression}=\frac{1}{\frac12\cos2\theta}=2\sec2\theta.
\displaystyle \therefore \lambda=2.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The value of }\cos^2\left(\frac{\pi}{6}+\theta\right)
\displaystyle -\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 \cos^2A-\sin^2B=\cos(A+B)\cos(A-B).
\displaystyle \text{Let }A=\frac{\pi}{6}+\theta,\qquad B=\frac{\pi}{6}-\theta.
\displaystyle \therefore A+B=\frac{\pi}{3},\qquad A-B=2\theta.
\displaystyle \therefore \cos^2\left(\frac{\pi}{6}+\theta\right)-\sin^2\left(\frac{\pi}{6}-\theta\right)
\displaystyle =\cos\frac{\pi}{3}\cos2\theta=\frac12\cos2\theta.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 35: }\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 \sin3\theta=3\sin\theta-4\sin^3\theta.
\displaystyle =\sin\theta(3-4\sin^2\theta)
\displaystyle =\sin\theta(1+2\cos2\theta).
\displaystyle \therefore \frac{\sin3\theta}{1+2\cos2\theta}=\sin\theta.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{The value of }2\sin^2B+4\cos(A+B)\sin A\sin B
\displaystyle +\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 4\cos(A+B)\sin A\sin B
\displaystyle =2\sin A\{\sin(A+2B)-\sin A\}.
\displaystyle =\cos2B-\cos(2A+2B)-1+\cos2A.
\displaystyle \therefore \text{Required expression}
\displaystyle =1-\cos2B+\cos2B-\cos(2A+2B)-1+\cos2A
\displaystyle \quad+\cos(2A+2B)
\displaystyle =\cos2A.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 37: }\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 2(\sin2\theta+2\cos^2\theta-1)
\displaystyle =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 \therefore \cos\theta-\sin\theta-\cos3\theta+\sin3\theta
\displaystyle =2\sin\theta(\sin2\theta+\cos2\theta).
\displaystyle \therefore \text{Required value}=\frac{2(\sin2\theta+\cos2\theta)}{2\sin\theta(\sin2\theta+\cos2\theta)}
\displaystyle =\frac{1}{\sin\theta}=\mathrm{cosec}\,\theta.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 38: }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 \therefore 2(1-2\sin^27\theta)\sin3\theta
\displaystyle =2\sin3\theta\cos14\theta.
\displaystyle =\sin(3\theta+14\theta)+\sin(3\theta-14\theta)
\displaystyle =\sin17\theta+\sin(-11\theta)
\displaystyle =\sin17\theta-\sin11\theta.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{If }\alpha\text{ and }\beta\text{ are acute angles satisfying}
\displaystyle \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
\displaystyle \text{(c) }\sqrt2\cot\beta\qquad\text{(d) }\frac{1}{\sqrt2}\cot\beta
\displaystyle \text{Answer:}
\displaystyle \text{Let }t=\tan\beta.\text{ Then }\cos2\beta=\frac{1-t^2}{1+t^2}.
\displaystyle \cos2\alpha=\frac{3\frac{1-t^2}{1+t^2}-1}{3-\frac{1-t^2}{1+t^2}}
\displaystyle =\frac{1-2t^2}{1+2t^2}.
\displaystyle \text{Also, }\cos2\alpha=\frac{1-\tan^2\alpha}{1+\tan^2\alpha}.
\displaystyle \therefore \tan^2\alpha=2t^2=2\tan^2\beta.
\displaystyle \text{Since }\alpha,\beta\text{ are acute, }\tan\alpha,\tan\beta>0.
\displaystyle \therefore \tan\alpha=\sqrt2\tan\beta.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 40: }\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 \therefore \tan^2\frac{\alpha}{2}=\frac{1+e}{1-e}\tan^2\frac{\theta}{2}.
\displaystyle =\frac{1+e}{1-e}\cdot\frac{1-\cos\theta}{1+\cos\theta}.
\displaystyle \cos\alpha=\frac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}.
\displaystyle =\frac{(1-e)(1+\cos\theta)-(1+e)(1-\cos\theta)}{(1-e)(1+\cos\theta)+(1+e)(1-\cos\theta)}
\displaystyle =\frac{2(\cos\theta-e)}{2(1-e\cos\theta)}
\displaystyle =\frac{\cos\theta-e}{1-e\cos\theta}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{If }(2^n+1)\theta=\pi,\text{ then}
\displaystyle 2^n\cos\theta\cos2\theta\cos2^2\theta\ldots\cos2^{n-1}\theta=
\displaystyle \text{(a) }-1\qquad\text{(b) }1\qquad\text{(c) }\frac12\qquad\text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin2^n\theta
\displaystyle =2^n\sin\theta\cos\theta\cos2\theta\cos2^2\theta\ldots\cos2^{n-1}\theta.
\displaystyle \text{Given }(2^n+1)\theta=\pi.
\displaystyle \therefore 2^n\theta=\pi-\theta.
\displaystyle \therefore \sin2^n\theta=\sin(\pi-\theta)=\sin\theta.
\displaystyle \therefore 2^n\sin\theta\cos\theta\cos2\theta\ldots\cos2^{n-1}\theta=\sin\theta.
\displaystyle \therefore 2^n\cos\theta\cos2\theta\ldots\cos2^{n-1}\theta=1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 42: }\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=\frac{2t}{1-t^2}.
\displaystyle \cos2\theta=\frac{1-t^2}{1+t^2}
\displaystyle \Rightarrow \sec2\theta=\frac{1+t^2}{1-t^2}.
\displaystyle \therefore \tan2\theta+\sec2\theta
\displaystyle =\frac{2t+1+t^2}{1-t^2}
\displaystyle =\frac{(1+t)^2}{(1-t)(1+t)}
\displaystyle =\frac{1+t}{1-t}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 43: }\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\sin^2\theta\cos^2\theta
\displaystyle =1-8\sin^2\theta\cos^2\theta.
\displaystyle =1-2(2\sin\theta\cos\theta)^2
\displaystyle =1-2\sin^22\theta
\displaystyle =\cos4\theta.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{The value of }\cos(36^\circ-A)\cos(36^\circ+A)
\displaystyle +\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 2\cos(36^\circ-A)\cos(36^\circ+A)
\displaystyle =\cos72^\circ+\cos2A.
\displaystyle 2\cos(54^\circ-A)\cos(54^\circ+A)
\displaystyle =\cos108^\circ+\cos2A.
\displaystyle \therefore 2(\text{Required expression})
\displaystyle =\cos72^\circ+\cos108^\circ+2\cos2A.
\displaystyle \cos108^\circ=\cos(180^\circ-72^\circ)=-\cos72^\circ.
\displaystyle \therefore 2(\text{Required expression})=2\cos2A.
\displaystyle \therefore \text{Required value}=\cos2A.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 45: }\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 \text{Let }t=\tan\theta.
\displaystyle \tan(60^\circ-\theta)=\frac{\sqrt3-t}{1+\sqrt3t},
\displaystyle \tan(60^\circ+\theta)=\frac{\sqrt3+t}{1-\sqrt3t}.
\displaystyle \therefore \tan\theta\tan(60^\circ-\theta)\tan(60^\circ+\theta)
\displaystyle =t\frac{(\sqrt3-t)(\sqrt3+t)}{(1+\sqrt3t)(1-\sqrt3t)}
\displaystyle =\frac{t(3-t^2)}{1-3t^2}.
\displaystyle =\frac{3t-t^3}{1-3t^2}=\tan3\theta.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{The value of }\tan\theta+\tan(60^\circ+\theta)
\displaystyle +\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 \text{Let }t=\tan\theta.
\displaystyle \tan(60^\circ+\theta)=\frac{\sqrt3+t}{1-\sqrt3t},
\displaystyle \tan(120^\circ+\theta)=\frac{t-\sqrt3}{1+\sqrt3t}.
\displaystyle \therefore \text{Required expression}
\displaystyle =t+\frac{\sqrt3+t}{1-\sqrt3t}+\frac{t-\sqrt3}{1+\sqrt3t}
\displaystyle =t+\frac{8t}{1-3t^2}
\displaystyle =\frac{9t-3t^3}{1-3t^2}
\displaystyle =3\frac{3t-t^3}{1-3t^2}
\displaystyle =3\tan3\theta.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 47: }\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
\displaystyle =2\cos4\alpha\sin\alpha.
\displaystyle \cos5\alpha+\cos3\alpha
\displaystyle =2\cos4\alpha\cos\alpha.
\displaystyle \therefore \cos5\alpha+2\cos4\alpha+\cos3\alpha
\displaystyle =2\cos4\alpha(1+\cos\alpha).
\displaystyle \therefore \text{Required value}
\displaystyle =\frac{2\cos4\alpha\sin\alpha}{2\cos4\alpha(1+\cos\alpha)}
\displaystyle =\frac{\sin\alpha}{1+\cos\alpha}
\displaystyle =\tan\frac{\alpha}{2}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 48: }\frac{\sin5\theta}{\sin\theta}\text{ is equal to}
\displaystyle \text{(a) }16\cos^4\theta-12\cos^2\theta+1
\displaystyle \text{(b) }16\cos^4\theta+12\cos^2\theta+1
\displaystyle \text{(c) }16\cos^4\theta-12\cos^2\theta-1
\displaystyle \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 \therefore \frac{\sin5\theta}{\sin\theta}
\displaystyle =16\sin^4\theta-20\sin^2\theta+5.
\displaystyle \text{Using }\sin^2\theta=1-\cos^2\theta,
\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{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{If }n=1,2,3,\ldots,\text{ then}
\displaystyle \cos\alpha\cos2\alpha\cos4\alpha\ldots\cos2^{n-1}\alpha\text{ is equal to}
\displaystyle \text{(a) }\frac{\sin2n\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 \sin2\alpha=2\sin\alpha\cos\alpha.
\displaystyle \sin4\alpha=2\sin2\alpha\cos2\alpha
\displaystyle =2^2\sin\alpha\cos\alpha\cos2\alpha.
\displaystyle \text{Continuing in this way,}
\displaystyle \sin2^n\alpha
\displaystyle =2^n\sin\alpha\cos\alpha\cos2\alpha\cos4\alpha\ldots\cos2^{n-1}\alpha.
\displaystyle \therefore \cos\alpha\cos2\alpha\cos4\alpha\ldots\cos2^{n-1}\alpha
\displaystyle =\frac{\sin2^n\alpha}{2^n\sin\alpha}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 50: }\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{1-\tan^2\theta}{1+\tan^2\theta}=\frac{b^2-a^2}{a^2+b^2}.
\displaystyle \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2ab}{a^2+b^2}.
\displaystyle \therefore b\cos2\theta+a\sin2\theta
\displaystyle =\frac{b(b^2-a^2)+2a^2b}{a^2+b^2}
\displaystyle =\frac{b(a^2+b^2)}{a^2+b^2}=b.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 51: }\text{If }\tan\alpha=\frac17,\ \tan\beta=\frac13,\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-\frac1{49}}{1+\frac1{49}}=\frac{24}{25}.
\displaystyle \tan\beta=\frac13.
\displaystyle \therefore \sin2\beta=\frac{2\tan\beta}{1+\tan^2\beta}=\frac35,
\displaystyle \cos2\beta=\frac{1-\tan^2\beta}{1+\tan^2\beta}=\frac45.
\displaystyle \therefore \sin4\beta=2\sin2\beta\cos2\beta
\displaystyle =2\left(\frac35\right)\left(\frac45\right)=\frac{24}{25}.
\displaystyle \therefore \cos2\alpha=\sin4\beta.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{The value of }\cos^2 48^\circ-\sin^2 12^\circ\text{ is}
\displaystyle \text{(a) }\frac{\sqrt5+1}{8}\qquad\text{(b) }\frac{\sqrt5-1}{8}
\displaystyle \text{(c) }\frac{\sqrt5+1}{5}\qquad\text{(d) }\frac{\sqrt5+1}{2\sqrt2}
\displaystyle \text{Answer:}
\displaystyle \cos^2 A-\sin^2 B=\cos(A+B)\cos(A-B).
\displaystyle \therefore \cos^2 48^\circ-\sin^2 12^\circ
\displaystyle =\cos60^\circ\cos36^\circ.
\displaystyle \cos60^\circ=\frac12,\qquad\cos36^\circ=\frac{\sqrt5+1}{4}.
\displaystyle \therefore \cos^2 48^\circ-\sin^2 12^\circ
\displaystyle =\frac12\left(\frac{\sqrt5+1}{4}\right)=\frac{\sqrt5+1}{8}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 1: }\text{If }(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2
\displaystyle =\lambda\cos^2\left(\frac{\alpha-\beta}{2}\right),\text{ write the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2
\displaystyle =\cos^2\alpha+\sin^2\alpha+\cos^2\beta+\sin^2\beta
\displaystyle \quad+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)
\displaystyle =2+2\cos(\alpha-\beta)
\displaystyle =2\left\{2\cos^2\left(\frac{\alpha-\beta}{2}\right)\right\}
\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right).
\displaystyle \therefore \lambda=4.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the value of }\sin\frac{\pi}{12}\sin\frac{5\pi}{12}.
\displaystyle \text{Answer:}
\displaystyle 2\sin A\sin B=\cos(A-B)-\cos(A+B).
\displaystyle \therefore \sin\frac{\pi}{12}\sin\frac{5\pi}{12}
\displaystyle =\frac12\left\{\cos\left(\frac{\pi}{12}-\frac{5\pi}{12}\right)-\cos\left(\frac{\pi}{12}+\frac{5\pi}{12}\right)\right\}
\displaystyle =\frac12\left\{\cos\left(-\frac{\pi}{3}\right)-\cos\frac{\pi}{2}\right\}
\displaystyle =\frac12\left(\frac12-0\right)=\frac14.
\displaystyle \therefore \sin\frac{\pi}{12}\sin\frac{5\pi}{12}=\frac14.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\sin A+\sin B=\alpha\text{ and }\cos A+\cos B=\beta,
\displaystyle \text{then write the value of }\tan\left(\frac{A+B}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)=\alpha.
\displaystyle \cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)=\beta.
\displaystyle \text{Dividing the two equations,}
\displaystyle \tan\left(\frac{A+B}{2}\right)=\frac{\alpha}{\beta}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If }\cos A=m\cos B,\text{ then write the value of}
\displaystyle \cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \cos A=m\cos B.
\displaystyle \cos A-\cos B=(m-1)\cos B.
\displaystyle \cos A+\cos B=(m+1)\cos B.
\displaystyle \therefore \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}.
\displaystyle \text{Using the transformation formulae,}
\displaystyle \cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right),
\displaystyle \cos A-\cos B=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right).
\displaystyle \therefore \frac{\cos A+\cos B}{\cos A-\cos B}
\displaystyle =-\cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right).
\displaystyle \therefore \cot\left(\frac{A+B}{2}\right)\cot\left(\frac{A-B}{2}\right)
\displaystyle =-\frac{m+1}{m-1}=\frac{m+1}{1-m}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Write the value of the expression }\frac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ}.
\displaystyle \text{Answer:}
\displaystyle 2\sin10^\circ\sin70^\circ=\cos60^\circ-\cos80^\circ.
\displaystyle \therefore 4\sin10^\circ\sin70^\circ=1-2\cos80^\circ.
\displaystyle \therefore 1-4\sin10^\circ\sin70^\circ=2\cos80^\circ.
\displaystyle \text{But }\cos80^\circ=\sin10^\circ.
\displaystyle \therefore \frac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ}
\displaystyle =\frac{2\sin10^\circ}{2\sin10^\circ}=1.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }A+B=\frac{\pi}{3}\text{ and }\cos A+\cos B=1,\text{ then find the value of}
\displaystyle \cos\left(\frac{A-B}{2}\right).
\displaystyle \text{Answer:}
\displaystyle \cos A+\cos B
\displaystyle =2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right).
\displaystyle \therefore 1=2\cos\frac{\pi}{6}\cos\left(\frac{A-B}{2}\right).
\displaystyle =\sqrt3\cos\left(\frac{A-B}{2}\right).
\displaystyle \therefore \cos\left(\frac{A-B}{2}\right)=\frac{1}{\sqrt3}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Write the value of }\sin12^\circ\sin48^\circ\sin54^\circ.
\displaystyle \text{Answer:}
\displaystyle 2\sin12^\circ\sin48^\circ=\cos36^\circ-\cos60^\circ.
\displaystyle \therefore \sin12^\circ\sin48^\circ\sin54^\circ
\displaystyle =\frac12(\cos36^\circ-\cos60^\circ)\sin54^\circ.
\displaystyle \text{Since }\sin54^\circ=\cos36^\circ,
\displaystyle =\frac12\left(\cos^2 36^\circ-\frac12\cos36^\circ\right).
\displaystyle \cos36^\circ=\frac{1+\sqrt5}{4}.
\displaystyle \therefore \cos^2 36^\circ-\frac12\cos36^\circ=\frac14.
\displaystyle \therefore \sin12^\circ\sin48^\circ\sin54^\circ=\frac18.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If }\sin2A=\lambda\sin2B,\text{ then write the value of }\frac{\lambda+1}{\lambda-1}.
\displaystyle \text{Answer:}
\displaystyle \sin2A=\lambda\sin2B\Rightarrow\lambda=\frac{\sin2A}{\sin2B}.
\displaystyle \therefore \frac{\lambda+1}{\lambda-1}
\displaystyle =\frac{\sin2A+\sin2B}{\sin2A-\sin2B}.
\displaystyle =\frac{2\sin(A+B)\cos(A-B)}{2\cos(A+B)\sin(A-B)}.
\displaystyle =\tan(A+B)\cot(A-B).
\displaystyle \therefore \frac{\lambda+1}{\lambda-1}=\tan(A+B)\cot(A-B).
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Write the value of }\frac{\sin A+\sin3A}{\cos A+\cos3A}.
\displaystyle \text{Answer:}
\displaystyle \sin A+\sin3A=2\sin2A\cos A.
\displaystyle \cos A+\cos3A=2\cos2A\cos A.
\displaystyle \therefore \frac{\sin A+\sin3A}{\cos A+\cos3A}
\displaystyle =\frac{2\sin2A\cos A}{2\cos2A\cos A}=\tan2A.
\displaystyle \therefore \text{Required value}=\tan2A.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }\cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D),
\displaystyle \text{then write the value }\tan A\tan B\tan C.
\displaystyle \text{Answer:}
\displaystyle \cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D).
\displaystyle \cos A\cos B(1-\tan A\tan B)\cos C\cos D(\tan C-\tan D)
\displaystyle =\cos A\cos B(1+\tan A\tan B)\cos C\cos D(\tan C+\tan D).
\displaystyle (1-\tan A\tan B)(\tan C-\tan D)
\displaystyle =(1+\tan A\tan B)(\tan C+\tan D).
\displaystyle -2\tan D-2\tan A\tan B\tan C=0.
\displaystyle \therefore \tan A\tan B\tan C=-\tan D.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If }\cos4x=1+k\sin^2x\cos^2x,\text{ then write the value of }k.
\displaystyle \text{Answer:}
\displaystyle \cos4x=1-2\sin^22x.
\displaystyle \sin2x=2\sin x\cos x
\displaystyle \Rightarrow \sin^22x=4\sin^2x\cos^2x.
\displaystyle \therefore \cos4x=1-8\sin^2x\cos^2x.
\displaystyle \text{Comparing with }\cos4x=1+k\sin^2x\cos^2x,
\displaystyle \therefore k=-8.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If }\tan\frac{x}{2}=\frac{m}{n},\text{ then write the value of }m\sin x+n\cos x.
\displaystyle \text{Answer:}
\displaystyle \tan\frac{x}{2}=\frac{m}{n}
\displaystyle \Rightarrow m\cos\frac{x}{2}=n\sin\frac{x}{2}.
\displaystyle m\sin x+n\cos x
\displaystyle =2m\sin\frac{x}{2}\cos\frac{x}{2}
\displaystyle \quad+n\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right).
\displaystyle =2n\sin^2\frac{x}{2}
\displaystyle \quad+n\left(\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\right)
\displaystyle =n\left(\sin^2\frac{x}{2}+\cos^2\frac{x}{2}\right)=n.
\displaystyle \therefore m\sin x+n\cos x=n.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{If }\frac{\pi}{2}<\theta<\frac{3\pi}{2},\text{ then write the value of}
\displaystyle \sqrt{\frac{1+\cos2\theta}{2}}.
\displaystyle \text{Answer:}
\displaystyle \frac{1+\cos2\theta}{2}=\cos^2\theta.
\displaystyle \therefore \sqrt{\frac{1+\cos2\theta}{2}}=\sqrt{\cos^2\theta}=|\cos\theta|.
\displaystyle \text{Since }\frac{\pi}{2}<\theta<\frac{3\pi}{2},\quad\cos\theta<0.
\displaystyle \therefore |\cos\theta|=-\cos\theta.
\displaystyle \therefore \text{Required value}=-\cos\theta.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }\frac{\pi}{2}<\theta<\pi,\text{ then write the value of}
\displaystyle \sqrt{2+\sqrt{2+2\cos2\theta}}\text{ in the simplest form.}
\displaystyle \text{Answer:}
\displaystyle 2+2\cos2\theta=4\cos^2\theta.
\displaystyle \therefore \sqrt{2+2\cos2\theta}=2|\cos\theta|.
\displaystyle \text{Since }\frac{\pi}{2}<\theta<\pi,\quad\cos\theta<0.
\displaystyle \therefore |\cos\theta|=-\cos\theta.
\displaystyle \therefore \sqrt{2+\sqrt{2+2\cos2\theta}}
\displaystyle =\sqrt{2-2\cos\theta}.
\displaystyle 1-\cos\theta=2\sin^2\frac{\theta}{2}.
\displaystyle \therefore \sqrt{2-2\cos\theta}
\displaystyle =\sqrt{4\sin^2\frac{\theta}{2}}=2\left|\sin\frac{\theta}{2}\right|.
\displaystyle \text{Since }\frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{2},\quad\sin\frac{\theta}{2}>0.
\displaystyle \therefore \text{Required value}=2\sin\frac{\theta}{2}.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{If }\frac{\pi}{2}<\theta<\pi,\text{ then write the value of}
\displaystyle \sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}.
\displaystyle \text{Answer:}
\displaystyle 1-\cos2\theta=2\sin^2\theta,\qquad 1+\cos2\theta=2\cos^2\theta.
\displaystyle \therefore \sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}
\displaystyle =\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}
\displaystyle =\sqrt{\tan^2\theta}=|\tan\theta|.
\displaystyle \text{Since }\frac{\pi}{2}<\theta<\pi,\quad\tan\theta<0.
\displaystyle \therefore |\tan\theta|=-\tan\theta.
\displaystyle \therefore \text{Required value}=-\tan\theta.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{If }\pi<\theta<\frac{3\pi}{2},\text{ then write the value of}
\displaystyle \sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}.
\displaystyle \text{Answer:}
\displaystyle 1-\cos2\theta=2\sin^2\theta,\qquad 1+\cos2\theta=2\cos^2\theta.
\displaystyle \therefore \sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}
\displaystyle =\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}
\displaystyle =\sqrt{\tan^2\theta}=|\tan\theta|.
\displaystyle \text{Since }\pi<\theta<\frac{3\pi}{2},\quad\tan\theta>0.
\displaystyle \therefore |\tan\theta|=\tan\theta.
\displaystyle \therefore \text{Required value}=\tan\theta.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In a right angled triangle }ABC,\text{ write the value of}
\displaystyle \sin^2A+\sin^2B+\sin^2C.
\displaystyle \text{Answer:}
\displaystyle \text{Let }C=90^\circ.\text{ Then }A+B=90^\circ.
\displaystyle \therefore \sin B=\cos A.
\displaystyle \sin^2A+\sin^2B+\sin^2C
\displaystyle =\sin^2A+\cos^2A+\sin^290^\circ
\displaystyle =1+1=2.
\displaystyle \therefore \text{Required value}=2.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Write the value of }\cos^276^\circ+\cos^216^\circ
\displaystyle -\cos76^\circ\cos16^\circ.
\displaystyle \text{Answer:}
\displaystyle \cos^276^\circ+\cos^216^\circ
\displaystyle =1+\cos(76^\circ+16^\circ)\cos(76^\circ-16^\circ)
\displaystyle =1+\cos92^\circ\cos60^\circ.
\displaystyle \cos76^\circ\cos16^\circ
\displaystyle =\frac12\{\cos92^\circ+\cos60^\circ\}.
\displaystyle \therefore \text{Required expression}
\displaystyle =1+\frac12\cos92^\circ-\frac12\cos92^\circ-\frac14
\displaystyle =\frac34.
\displaystyle \therefore \text{Required value}=\frac34.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\frac{\pi}{4}<\theta<\frac{\pi}{2},\text{ then write the value of}
\displaystyle \sqrt{1-\sin2\theta}.
\displaystyle \text{Answer:}
\displaystyle 1-\sin2\theta=1-2\sin\theta\cos\theta.
\displaystyle =\sin^2\theta+\cos^2\theta-2\sin\theta\cos\theta
\displaystyle =(\sin\theta-\cos\theta)^2.
\displaystyle \therefore \sqrt{1-\sin2\theta}=|\sin\theta-\cos\theta|.
\displaystyle \text{Since }\frac{\pi}{4}<\theta<\frac{\pi}{2},\quad\sin\theta>\cos\theta.
\displaystyle \therefore \text{Required value}=\sin\theta-\cos\theta.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Write the value of }\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}.
\displaystyle \text{Answer:}
\displaystyle \sin\frac{2\pi}{7}=2\sin\frac{\pi}{7}\cos\frac{\pi}{7}.
\displaystyle \sin\frac{4\pi}{7}=2\sin\frac{2\pi}{7}\cos\frac{2\pi}{7}.
\displaystyle \sin\frac{8\pi}{7}=2\sin\frac{4\pi}{7}\cos\frac{4\pi}{7}.
\displaystyle \therefore \sin\frac{8\pi}{7}
\displaystyle =8\sin\frac{\pi}{7}\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}.
\displaystyle \text{But }\sin\frac{8\pi}{7}=-\sin\frac{\pi}{7}.
\displaystyle \therefore 8\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}=-1.
\displaystyle \therefore \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}=-\frac18.
\displaystyle \therefore \text{Required value}=-\frac18.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{If }\tan A=\frac{1-\cos B}{\sin B},\text{ then find the value of }\tan2A.
\displaystyle \text{Answer:}
\displaystyle \tan A=\frac{1-\cos B}{\sin B}.
\displaystyle =\frac{2\sin^2\frac{B}{2}}{2\sin\frac{B}{2}\cos\frac{B}{2}}
\displaystyle =\tan\frac{B}{2}.
\displaystyle \therefore A=\frac{B}{2}\Rightarrow 2A=B.
\displaystyle \therefore \tan2A=\tan B.
\displaystyle \text{Also, }\tan2A=\frac{2\tan A}{1-\tan^2A}.
\displaystyle \therefore \text{Required value}=\tan B.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{If }\sin x+\cos x=a,\text{ find the value of }\sin^6x+\cos^6x.
\displaystyle \text{Answer:}
\displaystyle \sin x+\cos x=a.
\displaystyle \text{Squaring, }\sin^2x+\cos^2x+2\sin x\cos x=a^2.
\displaystyle \therefore 2\sin x\cos x=a^2-1.
\displaystyle \therefore \sin^2x\cos^2x=\frac{(a^2-1)^2}{4}.
\displaystyle \sin^6x+\cos^6x
\displaystyle =(\sin^2x+\cos^2x)^3
\displaystyle \quad-3\sin^2x\cos^2x(\sin^2x+\cos^2x)
\displaystyle =1-\frac{3(a^2-1)^2}{4}
\displaystyle =\frac{1+6a^2-3a^4}{4}.
\displaystyle \therefore \text{Required value}=\frac{1+6a^2-3a^4}{4}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\sin x+\cos x=a,\text{ find the value of }|\sin x-\cos x|.
\displaystyle \text{Answer:}
\displaystyle \sin x+\cos x=a.
\displaystyle \text{Squaring, }\sin^2x+\cos^2x+2\sin x\cos x=a^2.
\displaystyle \therefore 2\sin x\cos x=a^2-1.
\displaystyle (\sin x-\cos x)^2
\displaystyle =\sin^2x+\cos^2x-2\sin x\cos x
\displaystyle =1-(a^2-1)=2-a^2.
\displaystyle \therefore |\sin x-\cos x|=\sqrt{2-a^2}.
\displaystyle \therefore \text{Required value}=\sqrt{2-a^2}.
\displaystyle \\

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