\displaystyle \textbf{Question 1. }\text{The value of }\sin^275^\circ-\sin^215^\circ\text{ is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }\frac{\sqrt3}{2}\qquad \text{(c) }1\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sin^275^\circ-\sin^215^\circ
\displaystyle =\frac{1-\cos150^\circ}{2}-\frac{1-\cos30^\circ}{2}
\displaystyle =\frac{\cos30^\circ-\cos150^\circ}{2}
\displaystyle =\frac{\frac{\sqrt3}{2}-\left(-\frac{\sqrt3}{2}\right)}{2}=\frac{\sqrt3}{2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 2. }\text{If }A+B+C=180^\circ,\text{ then }\sec A(\cos B\cos C-\sin B\sin C)\text{ is equal to}
\displaystyle \text{(a) }0\qquad \text{(b) }-1\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos B\cos C-\sin B\sin C=\cos(B+C)
\displaystyle A+B+C=180^\circ\Rightarrow B+C=180^\circ-A
\displaystyle \cos(B+C)=\cos(180^\circ-A)=-\cos A
\displaystyle \therefore \sec A(\cos B\cos C-\sin B\sin C)=\sec A(-\cos A)=-1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ\text{ is equal to}
\displaystyle \text{(a) }\frac{\sqrt3}{4}\qquad \text{(b) }\frac{\sqrt3}{2}\qquad \text{(c) }\sqrt3\qquad \text{(d) }1
\displaystyle \text{Answer:}
\displaystyle \tan(20^\circ+40^\circ)=\tan60^\circ=\sqrt3
\displaystyle \frac{\tan20^\circ+\tan40^\circ}{1-\tan20^\circ\tan40^\circ}=\sqrt3
\displaystyle \tan20^\circ+\tan40^\circ=\sqrt3(1-\tan20^\circ\tan40^\circ)
\displaystyle \therefore \tan20^\circ+\tan40^\circ+\sqrt3\tan20^\circ\tan40^\circ=\sqrt3
\displaystyle \therefore \text{Correct option is (c).}
\\

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

\displaystyle \textbf{Question 5. }\text{If }3\sin\theta+4\cos\theta=5,\text{ then }4\sin\theta-3\cos\theta=
\displaystyle \text{(a) }0\qquad \text{(b) }5\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle (3\sin\theta+4\cos\theta)^2+(4\sin\theta-3\cos\theta)^2=25
\displaystyle 25+(4\sin\theta-3\cos\theta)^2=25
\displaystyle \therefore (4\sin\theta-3\cos\theta)^2=0
\displaystyle \therefore 4\sin\theta-3\cos\theta=0
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 6. }\text{If in }\Delta ABC,\ \tan A+\tan B+\tan C=6,\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=180^\circ
\displaystyle \therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C
\displaystyle \therefore \tan A\tan B\tan C=6
\displaystyle \therefore \cot A\cot B\cot C=\frac{1}{6}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 7. }\tan3A-\tan2A-\tan A\text{ is equal to}
\displaystyle \text{(a) }\tan3A\tan2A\tan A\qquad \text{(b) }-\tan3A\tan2A\tan A
\displaystyle \text{(c) }\tan A\tan2A-\tan2A\tan3A-\tan3A\tan A\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 3A=2A+A
\displaystyle \tan3A=\frac{\tan2A+\tan A}{1-\tan2A\tan A}
\displaystyle \tan3A-\tan2A-\tan A=\tan3A\tan2A\tan A
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 8. }\text{If }A+B+C=180^\circ,\text{ then }\frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}\text{ is equal to}
\displaystyle \text{(a) }\tan A\tan B\tan C\qquad \text{(b) }0\qquad \text{(c) }1\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle A+B+C=180^\circ
\displaystyle \therefore \tan A+\tan B+\tan C=\tan A\tan B\tan C
\displaystyle \therefore \frac{\tan A+\tan B+\tan C}{\tan A\tan B\tan C}=1
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 9. }\text{If }\cos P=\frac{1}{7}\text{ and }\cos Q=\frac{13}{14},\text{ where }P\text{ and }Q\text{ both are acute angles.} \\ \text{Then, the value of }P-Q\text{ is}
\displaystyle \text{(a) }30^\circ\qquad \text{(b) }60^\circ\qquad \text{(c) }45^\circ\qquad \text{(d) }75^\circ
\displaystyle \text{Answer:}
\displaystyle \sin P=\sqrt{1-\frac{1}{49}}=\frac{4\sqrt3}{7}
\displaystyle \sin Q=\sqrt{1-\frac{169}{196}}=\frac{3\sqrt3}{14}
\displaystyle \cos(P-Q)=\cos P\cos Q+\sin P\sin Q
\displaystyle =\frac{1}{7}\cdot\frac{13}{14}+\frac{4\sqrt3}{7}\cdot\frac{3\sqrt3}{14}
\displaystyle =\frac{13}{98}+\frac{36}{98}=\frac{1}{2}
\displaystyle \therefore P-Q=60^\circ
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 10. }\text{If }\cot(\alpha+\beta)=0,\text{ then }\sin(\alpha+2\beta)\text{ is equal to}
\displaystyle \text{(a) }\sin\alpha\qquad \text{(b) }\cos2\beta\qquad \text{(c) }\cos\alpha\qquad \text{(d) }\sin2\alpha
\displaystyle \text{Answer:}
\displaystyle \cot(\alpha+\beta)=0
\displaystyle \therefore \alpha+\beta=\frac{\pi}{2}
\displaystyle \sin(\alpha+2\beta)=\sin\left(\frac{\pi}{2}+\beta\right)
\displaystyle =\cos\beta
\displaystyle \text{Also, }\alpha+\beta=\frac{\pi}{2}\Rightarrow \cos\beta=\sin\alpha
\displaystyle \therefore \sin(\alpha+2\beta)=\sin\alpha
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 11. }\frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}\text{ is equal to}
\displaystyle \text{(a) }\tan55^\circ\qquad \text{(b) }\cot55^\circ\qquad \text{(c) }-\tan35^\circ\qquad \text{(d) }-\cot35^\circ
\displaystyle \text{Answer:}
\displaystyle \frac{\cos10^\circ+\sin10^\circ}{\cos10^\circ-\sin10^\circ}=\frac{1+\tan10^\circ}{1-\tan10^\circ}
\displaystyle =\tan(45^\circ+10^\circ)
\displaystyle =\tan55^\circ
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 12. }\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 A=\frac{\pi}{6}+\theta,\quad B=\frac{\pi}{6}-\theta
\displaystyle \therefore \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}=\frac{1}{2}\cos2\theta
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 13. }\text{If }\tan\theta_1\tan\theta_2=k,\text{ then }\frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=
\displaystyle \text{(a) }\frac{1+k}{1-k}\qquad \text{(b) }\frac{1-k}{1+k}\qquad \text{(c) }\frac{k+1}{k-1}\qquad \text{(d) }\frac{k-1}{k+1}
\displaystyle \text{Answer:}
\displaystyle \frac{\cos(\theta_1-\theta_2)}{\cos(\theta_1+\theta_2)}=\frac{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2}
\displaystyle =\frac{1+\tan\theta_1\tan\theta_2}{1-\tan\theta_1\tan\theta_2}
\displaystyle =\frac{1+k}{1-k}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 14. }\text{If }\sin(\pi\cos\theta)=\cos(\pi\sin\theta),\text{ then }\sin2\theta=
\displaystyle \text{(a) }\pm\frac{3}{4}\qquad \text{(b) }\pm\frac{4}{3}\qquad \text{(c) }\pm\frac{1}{3}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \cos(\pi\sin\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right)
\displaystyle \sin(\pi\cos\theta)=\sin\left(\frac{\pi}{2}-\pi\sin\theta\right)
\displaystyle \pi\cos\theta=\frac{\pi}{2}-\pi\sin\theta
\displaystyle \therefore \sin\theta+\cos\theta=\frac{1}{2}
\displaystyle \therefore (\sin\theta+\cos\theta)^2=\frac{1}{4}
\displaystyle 1+\sin2\theta=\frac{1}{4}
\displaystyle \therefore \sin2\theta=-\frac{3}{4}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 15. }\text{If }\tan\theta=\frac{1}{2}\text{ and }\tan\phi=\frac{1}{3},\text{ then the value of }\theta+\phi\text{ is}
\displaystyle \text{(a) }\frac{\pi}{6}\qquad \text{(b) }\pi\qquad \text{(c) }0\qquad \text{(d) }\frac{\pi}{4}
\displaystyle \text{Answer:}
\displaystyle \tan(\theta+\phi)=\frac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}
\displaystyle =\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}
\displaystyle =\frac{\frac{5}{6}}{\frac{5}{6}}=1
\displaystyle \therefore \theta+\phi=\frac{\pi}{4}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 16. }\text{The value of } \\ \cos(36^\circ-A)\cos(36^\circ+A)+\cos(54^\circ+A)\cos(54^\circ-A)\text{ is}
\displaystyle \text{(a) }\sin2A\qquad \text{(b) }\cos2A\qquad \text{(c) }\cos3A\qquad \text{(d) }\sin3A
\displaystyle \text{Answer:}
\displaystyle \cos(36^\circ-A)\cos(36^\circ+A)=\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 (b).}
\\

\displaystyle \textbf{Question 17. }\text{If }\tan\left(\frac{\pi}{4}+\theta\right)+\tan\left(\frac{\pi}{4}-\theta\right)=a,\text{ then }\tan^2\left(\frac{\pi}{4}+\theta\right)+\tan^2\left(\frac{\pi}{4}-\theta\right)=
\displaystyle \text{(a) }a^2+1\qquad \text{(b) }a^2+2\qquad \text{(c) }a^2-2\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \tan\left(\frac{\pi}{4}+\theta\right)\tan\left(\frac{\pi}{4}-\theta\right)=1
\displaystyle \text{Let }p=\tan\left(\frac{\pi}{4}+\theta\right),\ q=\tan\left(\frac{\pi}{4}-\theta\right)
\displaystyle p+q=a,\quad pq=1
\displaystyle p^2+q^2=(p+q)^2-2pq
\displaystyle =a^2-2
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 18. }\text{If }\tan(A-B)=1,\ \sec(A+B)=\frac{2}{\sqrt3},\text{ then the smallest positive value of }B\text{ is}
\displaystyle \text{(a) }\frac{25\pi}{24}\qquad \text{(b) }\frac{19\pi}{24}\qquad \text{(c) }\frac{13\pi}{24}\qquad \text{(d) }\frac{11\pi}{24}
\displaystyle \text{Answer:}
\displaystyle \tan(A-B)=1
\displaystyle \therefore A-B=\frac{\pi}{4}\qquad \text{(taking smallest positive value)}
\displaystyle \sec(A+B)=\frac{2}{\sqrt{3}}
\displaystyle \therefore \cos(A+B)=\frac{\sqrt{3}}{2}
\displaystyle \therefore A+B=\frac{\pi}{6}\qquad \text{or}\qquad \frac{11\pi}{6}
\displaystyle \text{Using }A+B=\frac{\pi}{6}
\displaystyle 2A=\frac{\pi}{6}+\frac{\pi}{4}=\frac{5\pi}{12}
\displaystyle A=\frac{5\pi}{24}
\displaystyle B=\frac{\pi}{6}-\frac{5\pi}{24}=-\frac{\pi}{24}\qquad \text{(not positive)}
\displaystyle \text{Using }A+B=\frac{11\pi}{6}
\displaystyle 2A=\frac{11\pi}{6}+\frac{\pi}{4}=\frac{25\pi}{12}
\displaystyle A=\frac{25\pi}{24}
\displaystyle B=\frac{11\pi}{6}-\frac{25\pi}{24}=\frac{19\pi}{24}
\displaystyle \therefore\text{Smallest positive value of }B=\frac{19\pi}{24}
\displaystyle \therefore\text{Correct option is (b).}
\\

\displaystyle \textbf{Question 19. }\text{If }A-B=\frac{\pi}{4},\text{ then }(1+\tan A)(1-\tan B)\text{ is equal to}
\displaystyle \text{(a) }2\qquad \text{(b) }1\qquad \text{(c) }0\qquad \text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \tan(A-B)=\tan\frac{\pi}{4}=1
\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}=1
\displaystyle \tan A-\tan B=1+\tan A\tan B
\displaystyle 1+\tan A-\tan B-\tan A\tan B=2
\displaystyle (1+\tan A)(1-\tan B)=2
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 20. }\text{The maximum value of }\sin^2(120^\circ+\theta)+\sin^2(120^\circ-\theta)\text{ is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }\frac{3}{2}\qquad \text{(c) }\frac{1}{4}\qquad \text{(d) }\frac{3}{4}
\displaystyle \text{Answer:}
\displaystyle \sin^2A+\sin^2B=1-\frac{\cos2A+\cos2B}{2}
\displaystyle A=120^\circ+\theta,\quad B=120^\circ-\theta
\displaystyle \cos2A+\cos2B=\cos(240^\circ+2\theta)+\cos(240^\circ-2\theta)
\displaystyle =2\cos240^\circ\cos2\theta=-\cos2\theta
\displaystyle \therefore \sin^2(120^\circ+\theta)+\sin^2(120^\circ-\theta)=1+\frac{\cos2\theta}{2}
\displaystyle \therefore \text{Maximum value}=\frac{3}{2}
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 21. }\text{If }\cos(A-B)=\frac{3}{5}\text{ and }\tan A\tan B=2,\text{ then}
\displaystyle \text{(a) }\cos A\cos B=\frac{1}{5}\qquad \text{(b) }\cos A\cos B=-\frac{1}{5}
\displaystyle \text{(c) }\sin A\sin B=-\frac{1}{5}\qquad \text{(d) }\sin A\sin B=-\frac{1}{5}
\displaystyle \text{Answer:}
\displaystyle \tan A\tan B=2
\displaystyle \therefore \sin A\sin B=2\cos A\cos B
\displaystyle \cos(A-B)=\cos A\cos B+\sin A\sin B
\displaystyle \frac{3}{5}=3\cos A\cos B
\displaystyle \therefore \cos A\cos B=\frac{1}{5}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 22. }\text{If }\tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=2k,\text{ then }k=
\displaystyle \text{(a) }-1\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }-\frac{1}{2}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle 69^\circ+66^\circ=135^\circ
\displaystyle \tan135^\circ=-1
\displaystyle \frac{\tan69^\circ+\tan66^\circ}{1-\tan69^\circ\tan66^\circ}=-1
\displaystyle \tan69^\circ+\tan66^\circ=-1+\tan69^\circ\tan66^\circ
\displaystyle \tan69^\circ+\tan66^\circ-\tan69^\circ\tan66^\circ=-1
\displaystyle \therefore 2k=-1
\displaystyle \therefore k=-\frac{1}{2}
\displaystyle \therefore \text{Correct option is (c).}
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\displaystyle \textbf{Question 23. }\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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