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

\displaystyle \textbf{Question 2. }\sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ=
\displaystyle \text{(a) }0\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin163^\circ=\sin17^\circ,\quad \cos347^\circ=\cos13^\circ
\displaystyle \sin73^\circ=\cos17^\circ,\quad \sin167^\circ=\sin13^\circ
\displaystyle \therefore \sin163^\circ\cos347^\circ+\sin73^\circ\sin167^\circ
\displaystyle =\sin17^\circ\cos13^\circ+\cos17^\circ\sin13^\circ
\displaystyle =\sin30^\circ=\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\text{If }\sin2\theta+\sin2\phi=\frac{1}{2}\text{ and }\cos2\theta+\cos2\phi=\frac{3}{2},\text{ then } \\ \cos^2(\theta-\phi)=
\displaystyle \text{(a) }\frac{3}{8}\qquad \text{(b) }\frac{5}{8}\qquad \text{(c) }\frac{3}{4}\qquad \text{(d) }\frac{5}{4}
\displaystyle \text{Answer:}
\displaystyle (\sin2\theta+\sin2\phi)^2+(\cos2\theta+\cos2\phi)^2
\displaystyle =2+2\cos(2\theta-2\phi)
\displaystyle \left(\frac{1}{2}\right)^2+\left(\frac{3}{2}\right)^2=2+2\cos2(\theta-\phi)
\displaystyle \frac{5}{2}=2+2\cos2(\theta-\phi)
\displaystyle \cos2(\theta-\phi)=\frac{1}{4}
\displaystyle \cos^2(\theta-\phi)=\frac{1+\cos2(\theta-\phi)}{2}=\frac{5}{8}
\displaystyle \therefore \text{Correct option is (b).}
\\

\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) }\frac{3}{2}
\displaystyle \text{Answer:}
\displaystyle \cos52^\circ+\cos68^\circ=2\cos60^\circ\cos8^\circ=\cos8^\circ
\displaystyle \cos172^\circ=-\cos8^\circ
\displaystyle \therefore \cos52^\circ+\cos68^\circ+\cos172^\circ=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 5. }\text{The value of }\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ\text{ is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }-\frac{1}{2}\qquad \text{(c) }-1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sin78^\circ-\sin42^\circ=2\cos60^\circ\sin18^\circ=\sin18^\circ
\displaystyle \sin66^\circ-\sin6^\circ=2\cos36^\circ\sin30^\circ=\cos36^\circ
\displaystyle \therefore \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ
\displaystyle =\sin18^\circ-\cos36^\circ
\displaystyle =\sin18^\circ-\sin54^\circ=-\frac{1}{2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 6. }\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 \therefore \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 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 \cos155^\circ=\cos(180^\circ-25^\circ)=-\cos25^\circ
\displaystyle \therefore S=\cos35^\circ+\cos85^\circ-\cos25^\circ
\displaystyle \cos85^\circ+\cos35^\circ=2\cos60^\circ\cos25^\circ
\displaystyle \cos85^\circ+\cos35^\circ=\cos25^\circ
\displaystyle \therefore S=\cos25^\circ-\cos25^\circ
\displaystyle S=0
\displaystyle \therefore\text{Correct option is (a).}
\\

\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) }\frac{1}{2}\qquad \text{(d) }2
\displaystyle \text{Answer:}
\displaystyle \sin50^\circ+\sin10^\circ=2\sin30^\circ\cos20^\circ=\cos20^\circ
\displaystyle \sin70^\circ=\cos20^\circ
\displaystyle \therefore \sin50^\circ-\sin70^\circ+\sin10^\circ=0
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 \sin47^\circ+\sin61^\circ-\sin11^\circ-\sin25^\circ
\displaystyle =2\sin18^\circ(\cos29^\circ+\cos43^\circ)
\displaystyle =4\sin18^\circ\cos36^\circ\cos7^\circ
\displaystyle =2\sin36^\circ\cos7^\circ
\displaystyle =\sin43^\circ+\sin29^\circ
\displaystyle =\cos47^\circ+\cos61^\circ
\displaystyle =\cos7^\circ
\displaystyle \therefore \text{Correct option is (d).}
\\

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

\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=2\cos\frac{A+C}{2}\sin\frac{A-C}{2}
\displaystyle =2\cos B\sin\frac{A-C}{2}
\displaystyle \cos C-\cos A=2\sin\frac{A+C}{2}\sin\frac{A-C}{2}
\displaystyle =2\sin B\sin\frac{A-C}{2}
\displaystyle \therefore \frac{\sin A-\sin C}{\cos C-\cos A}=\cot B
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 12. }\text{If }\sin(B+C-A),\ \sin(C+A-B),\ \sin(A+B-C)\text{ are in A.P., then } \\ \cot A,\cot B,\cot C\text{ are in}
\displaystyle \text{(a) GP}\qquad \text{(b) HP}\qquad \text{(c) AP}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle A+B+C=\pi
\displaystyle \therefore B+C-A=\pi-2A
\displaystyle C+A-B=\pi-2B
\displaystyle A+B-C=\pi-2C
\displaystyle \therefore \sin(B+C-A)=\sin2A
\displaystyle \sin(C+A-B)=\sin2B
\displaystyle \sin(A+B-C)=\sin2C
\displaystyle \text{Since these are in A.P.,}
\displaystyle \sin2A+\sin2C=2\sin2B
\displaystyle 2\sin(A+C)\cos(A-C)=4\sin B\cos B
\displaystyle 2\sin B\cos(A-C)=4\sin B\cos B
\displaystyle \cos(A-C)=2\cos B
\displaystyle \cos A\cos C+\sin A\sin C=2{\sin A\sin C-\cos A\cos C}
\displaystyle 3\cos A\cos C=\sin A\sin C
\displaystyle \cot A\cot C=\frac{1}{3}
\displaystyle \text{This condition implies that }\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{Correct option is (b).}
\\

\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 2\sin\frac{x+y}{2}\cos\frac{x-y}{2}
\displaystyle =\sqrt3\left(-2\sin\frac{x+y}{2}\sin\frac{y-x}{2}\right)
\displaystyle \cos\frac{x-y}{2}=\sqrt3\sin\frac{x-y}{2}
\displaystyle \tan\frac{x-y}{2}=\frac{1}{\sqrt3}
\displaystyle \therefore \frac{x-y}{2}=\frac{\pi}{6}
\displaystyle x-y=\frac{\pi}{3}
\displaystyle \sin3x+\sin3y=2\sin\frac{3x+3y}{2}\cos\frac{3x-3y}{2}
\displaystyle =2\sin\frac{3(x+y)}{2}\cos\frac{3\pi}{2}=0
\displaystyle \therefore \text{Correct option is (b).}
\\

\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 =1
\displaystyle \therefore \alpha+\beta=\frac{\pi}{4}
\displaystyle \therefore \text{Correct option is (d).}
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