\displaystyle \textbf{Question 1. }\text{If }\tan\theta=x-\frac{1}{4x},\text{ then }\sec\theta-\tan\theta\text{ is equal to}
\displaystyle \text{(a) }-2x,\frac{1}{2x}\qquad \text{(b) }-\frac{1}{2x},2x\qquad \text{(c) }2x\qquad \text{(d) }2x,\frac{1}{2x}
\displaystyle \text{Answer:}
\displaystyle \tan\theta=x-\frac{1}{4x}
\displaystyle \tan^2\theta=\left(x-\frac{1}{4x}\right)^2
\displaystyle \tan^2\theta=x^2-\frac{1}{2}+\frac{1}{16x^2}
\displaystyle \sec^2\theta=1+\tan^2\theta
\displaystyle \sec^2\theta=1+x^2-\frac{1}{2}+\frac{1}{16x^2}
\displaystyle \sec^2\theta=x^2+\frac{1}{2}+\frac{1}{16x^2}
\displaystyle \sec^2\theta=\left(x+\frac{1}{4x}\right)^2
\displaystyle \therefore\sec\theta=x+\frac{1}{4x}
\displaystyle \sec\theta-\tan\theta=\left(x+\frac{1}{4x}\right)-\left(x-\frac{1}{4x}\right)
\displaystyle \sec\theta-\tan\theta=\frac{1}{2x}
\displaystyle \sec\theta+\tan\theta=\left(x+\frac{1}{4x}\right)+\left(x-\frac{1}{4x}\right)
\displaystyle \sec\theta+\tan\theta=2x
\displaystyle \therefore\text{Correct option is (a).}
\\

\displaystyle \textbf{Question 2. }\text{If }\sec\theta=x+\frac{1}{4x},\text{ then }\sec\theta+\tan\theta=
\displaystyle \text{(a) }x,\frac{1}{x}\qquad \text{(b) }2x,\frac{1}{2x}\qquad \text{(c) }-2x,\frac{1}{2x}\qquad \text{(d) }-\frac{1}{x},x
\displaystyle \text{Answer:}
\displaystyle \sec\theta+\tan\theta=t,\text{ then }\sec\theta-\tan\theta=\frac{1}{t}
\displaystyle \therefore 2\sec\theta=t+\frac{1}{t}
\displaystyle 2\left(x+\frac{1}{4x}\right)=t+\frac{1}{t}
\displaystyle \frac{4x^2+1}{2x}=t+\frac{1}{t}
\displaystyle \therefore t=2x\text{ or }t=\frac{1}{2x}
\displaystyle \therefore \sec\theta+\tan\theta=2x\text{ or }\frac{1}{2x}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 3. }\text{If }\frac{\pi}{2}<\theta<\frac{3\pi}{2},\text{ then }\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}\text{ is equal to}
\displaystyle \text{(a) }\sec\theta-\tan\theta\qquad \text{(b) }\sec\theta+\tan\theta\qquad \text{(c) }\tan\theta-\sec\theta\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\frac{(1-\sin\theta)^2}{1-\sin^2\theta}}
\displaystyle =\sqrt{\frac{(1-\sin\theta)^2}{\cos^2\theta}}
\displaystyle =\left|\frac{1-\sin\theta}{\cos\theta}\right|
\displaystyle =|\sec\theta-\tan\theta|
\displaystyle \frac{\pi}{2}<\theta<\frac{3\pi}{2}\Rightarrow \cos\theta<0
\displaystyle \therefore \frac{1-\sin\theta}{\cos\theta}<0
\displaystyle \therefore \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\tan\theta-\sec\theta
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 4. }\text{If }\pi<\theta<2\pi,\text{ then }\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\text{ is equal to}
\displaystyle \text{(a) }\mathrm{cosec}\ \theta+\cot\theta\qquad \text{(b) }\mathrm{cosec}\ \theta-\cot\theta
\displaystyle \text{(c) }-\mathrm{cosec}\ \theta+\cot\theta\qquad \text{(d) }-\mathrm{cosec}\ \theta-\cot\theta
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}=\sqrt{\frac{(1+\cos\theta)^2}{1-\cos^2\theta}}
\displaystyle =\sqrt{\frac{(1+\cos\theta)^2}{\sin^2\theta}}
\displaystyle =\left|\frac{1+\cos\theta}{\sin\theta}\right|
\displaystyle =|\mathrm{cosec}\ \theta+\cot\theta|
\displaystyle \pi<\theta<2\pi\Rightarrow \sin\theta<0
\displaystyle \therefore \frac{1+\cos\theta}{\sin\theta}<0
\displaystyle \therefore \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}=-\mathrm{cosec}\ \theta-\cot\theta
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 5. }\text{If }0<\theta<\frac{\pi}{2},\text{ and if }\frac{y+1}{1-y}=\sqrt{\frac{1+\sin\theta}{1-\sin\theta}},\text{ then }y\text{ is equal to}
\displaystyle \text{(a) }\cot\frac{\theta}{2}\qquad \text{(b) }\tan\frac{\theta}{2}\qquad \text{(c) }\cot\frac{\theta}{2}+\tan\frac{\theta}{2}\qquad \text{(d) }\cot\frac{\theta}{2}-\tan\frac{\theta}{2}
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=\frac{1+\sin\theta}{\cos\theta}
\displaystyle =\sec\theta+\tan\theta
\displaystyle =\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}
\displaystyle \therefore \frac{y+1}{1-y}=\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}
\displaystyle \therefore y=\tan\frac{\theta}{2}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 6. }\text{If }\frac{\pi}{2}<\theta<\pi,\text{ then }\sqrt{\frac{1-\sin\theta}{1+\sin\theta}}+\sqrt{\frac{1+\sin\theta}{1-\sin\theta}}\text{ is equal to}
\displaystyle \text{(a) }2\sec\theta\qquad \text{(b) }-2\sec\theta\qquad \text{(c) }\sec\theta\qquad \text{(d) }-\sec\theta
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\sin\theta}{1+\sin\theta}}=\tan\theta-\sec\theta
\displaystyle \sqrt{\frac{1+\sin\theta}{1-\sin\theta}}=-(\sec\theta+\tan\theta)
\displaystyle \therefore \text{Sum}=\tan\theta-\sec\theta-\sec\theta-\tan\theta
\displaystyle =-2\sec\theta
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 7. }\text{If }x=r\sin\theta\cos\phi,\ y=r\sin\theta\sin\phi\text{ and }z=r\cos\theta,\text{ then }x^2+y^2+z^2\text{ is independent of}
\displaystyle \text{(a) }\theta,\phi\qquad \text{(b) }r,\theta\qquad \text{(c) }r,\phi\qquad \text{(d) }r
\displaystyle \text{Answer:}
\displaystyle x^2+y^2+z^2=r^2\sin^2\theta\cos^2\phi+r^2\sin^2\theta\sin^2\phi+r^2\cos^2\theta
\displaystyle =r^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+r^2\cos^2\theta
\displaystyle =r^2(\sin^2\theta+\cos^2\theta)=r^2
\displaystyle \therefore x^2+y^2+z^2\text{ is independent of }\theta\text{ and }\phi
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 8. }\text{If }\tan\theta+\sec\theta=\sqrt3,\ 0<\theta<\pi,\text{ then }\theta\text{ is equal to}
\displaystyle \text{(a) }\frac{5\pi}{6}\qquad \text{(b) }\frac{2\pi}{3}\qquad \text{(c) }\frac{\pi}{6}\qquad \text{(d) }\frac{\pi}{3}
\displaystyle \text{Answer:}
\displaystyle \tan\theta+\sec\theta=\sqrt3
\displaystyle \sec\theta-\tan\theta=\frac{1}{\sqrt3}
\displaystyle 2\sec\theta=\sqrt3+\frac{1}{\sqrt3}=\frac{4}{\sqrt3}
\displaystyle \sec\theta=\frac{2}{\sqrt3}
\displaystyle \cos\theta=\frac{\sqrt3}{2}
\displaystyle 0<\theta<\pi\Rightarrow \theta=\frac{\pi}{6}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 9. }\text{If }\tan\theta=-\frac{1}{\sqrt5}\text{ and }\theta\text{ lies in the IV quadrant, then the value of }\cos\theta\text{ is}
\displaystyle \text{(a) }\frac{\sqrt5}{\sqrt6}\qquad \text{(b) }\frac{2}{\sqrt6}\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) }\frac{1}{\sqrt6}
\displaystyle \text{Answer:}
\displaystyle \tan\theta=-\frac{1}{\sqrt5}
\displaystyle \sec^2\theta=1+\tan^2\theta=1+\frac{1}{5}=\frac{6}{5}
\displaystyle \therefore \cos^2\theta=\frac{5}{6}
\displaystyle \theta\text{ lies in IV quadrant, so }\cos\theta>0
\displaystyle \therefore \cos\theta=\frac{\sqrt5}{\sqrt6}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 10. }\text{If }\frac{3\pi}{4}<\alpha<\pi,\text{ then }\sqrt{2\cot\alpha+\frac{1}{\sin^2\alpha}}\text{ is equal to}
\displaystyle \text{(a) }1-\cot\alpha\qquad \text{(b) }1+\cot\alpha\qquad \text{(c) }-1+\cot\alpha\qquad \text{(d) }-1-\cot\alpha
\displaystyle \text{Answer:}
\displaystyle 2\cot\alpha+\frac{1}{\sin^2\alpha}=2\cot\alpha+\mathrm{cosec}^2\alpha
\displaystyle =2\cot\alpha+1+\cot^2\alpha
\displaystyle =(1+\cot\alpha)^2
\displaystyle \frac{3\pi}{4}<\alpha<\pi\Rightarrow -1<\cot\alpha<0
\displaystyle \therefore 1+\cot\alpha>0
\displaystyle \therefore \sqrt{2\cot\alpha+\frac{1}{\sin^2\alpha}}=1+\cot\alpha
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 11. }\sin^6A+\cos^6A+3\sin^2A\cos^2A=
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }2\qquad \text{(d) }3
\displaystyle \text{Answer:}
\displaystyle \sin^6A+\cos^6A=(\sin^2A+\cos^2A)^3-3\sin^2A\cos^2A(\sin^2A+\cos^2A)
\displaystyle =1-3\sin^2A\cos^2A
\displaystyle \therefore \sin^6A+\cos^6A+3\sin^2A\cos^2A=1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 12. }\text{If }\mathrm{cosec}\ \theta-\cot\theta=\frac{1}{2},\ 0<\theta<\frac{\pi}{2},\text{ then }\cos\theta\text{ is equal to}
\displaystyle \text{(a) }\frac{5}{3}\qquad \text{(b) }\frac{3}{5}\qquad \text{(c) }-\frac{3}{5}\qquad \text{(d) }-\frac{5}{3}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\ \theta-\cot\theta=\frac{1}{2}
\displaystyle \mathrm{cosec}\ \theta+\cot\theta=2
\displaystyle 2\mathrm{cosec}\ \theta=\frac{5}{2}
\displaystyle \mathrm{cosec}\ \theta=\frac{5}{4}
\displaystyle \sin\theta=\frac{4}{5}
\displaystyle \cos\theta=\frac{3}{5}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 13. }\text{If }\mathrm{cosec}\ \theta+\cot\theta=\frac{11}{2},\text{ then }\tan\theta=
\displaystyle \text{(a) }\frac{21}{22}\qquad \text{(b) }\frac{15}{16}\qquad \text{(c) }\frac{44}{117}\qquad \text{(d) }\frac{117}{44}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\ \theta+\cot\theta=\frac{11}{2}
\displaystyle \mathrm{cosec}\ \theta-\cot\theta=\frac{2}{11}
\displaystyle 2\mathrm{cosec}\ \theta=\frac{11}{2}+\frac{2}{11}=\frac{125}{22}
\displaystyle \mathrm{cosec}\ \theta=\frac{125}{44}
\displaystyle 2\cot\theta=\frac{11}{2}-\frac{2}{11}=\frac{117}{22}
\displaystyle \cot\theta=\frac{117}{44}
\displaystyle \therefore \tan\theta=\frac{44}{117}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 14. }\sec^2\theta=\frac{4xy}{(x+y)^2}\text{ is true if and only if}
\displaystyle \text{(a) }x+y\neq0\qquad \text{(b) }x=y,\ x\neq0\qquad \text{(c) }x=y\qquad \text{(d) }x\neq0,\ y\neq0
\displaystyle \text{Answer:}
\displaystyle \sec^2\theta\geq1
\displaystyle \frac{4xy}{(x+y)^2}\leq1\text{ by }(x-y)^2\geq0
\displaystyle \therefore \frac{4xy}{(x+y)^2}=1\text{ only when }x=y
\displaystyle \text{Also, denominator should not be zero, so }x=y\neq0
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 15. }\text{If }\theta\text{ is an acute angle and }\tan\theta=\frac{1}{\sqrt7},\text{ then the value of }\frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}\text{ is}
\displaystyle \text{(a) }\frac{3}{4}\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }2\qquad \text{(d) }\frac{5}{4}
\displaystyle \text{Answer:}
\displaystyle \tan\theta=\frac{1}{\sqrt7}
\displaystyle \sin\theta=\frac{1}{\sqrt8},\quad \cos\theta=\frac{\sqrt7}{\sqrt8}
\displaystyle \mathrm{cosec}^2\theta=8,\quad \sec^2\theta=\frac{8}{7}
\displaystyle \frac{\mathrm{cosec}^2\theta-\sec^2\theta}{\mathrm{cosec}^2\theta+\sec^2\theta}=\frac{8-\frac{8}{7}}{8+\frac{8}{7}}
\displaystyle =\frac{6}{8}=\frac{3}{4}
\displaystyle \therefore \text{Correct option is (a).}
\\

\displaystyle \textbf{Question 16. }\text{The value of }\sin^25^\circ+\sin^210^\circ+\sin^215^\circ+\cdots+\sin^285^\circ+\sin^290^\circ\text{ is}
\displaystyle \text{(a) }7\qquad \text{(b) }8\qquad \text{(c) }9.5\qquad \text{(d) }10
\displaystyle \text{Answer:}
\displaystyle S=\sin^2 5^\circ+\sin^2 10^\circ+\sin^2 15^\circ+\cdots+\sin^2 85^\circ+\sin^2 90^\circ
\displaystyle S=(\sin^2 5^\circ+\sin^2 85^\circ)+(\sin^2 10^\circ+\sin^2 80^\circ)+\cdots+(\sin^2 40^\circ+\sin^2 50^\circ)+\sin^2 45^\circ+\sin^2 90^\circ
\displaystyle \sin^2\theta+\sin^2(90^\circ-\theta)=\sin^2\theta+\cos^2\theta=1
\displaystyle \therefore S=8+\sin^2 45^\circ+\sin^2 90^\circ
\displaystyle S=8+\frac{1}{2}+1
\displaystyle S=\frac{19}{2}=9.5
\displaystyle \therefore\text{Correct option is (c).}
\\

\displaystyle \textbf{Question 17. }\sin^2\frac{\pi}{18}+\sin^2\frac{2\pi}{9}+\sin^2\frac{7\pi}{18}+\sin^2\frac{4\pi}{9}=
\displaystyle \text{(a) }1\qquad \text{(b) }4\qquad \text{(c) }2\qquad \text{(d) }0
\displaystyle \text{Answer:}
\displaystyle \sin^2\frac{\pi}{18}+\sin^2\frac{4\pi}{9}=1
\displaystyle \sin^2\frac{2\pi}{9}+\sin^2\frac{7\pi}{18}=1
\displaystyle \therefore \text{Required value}=2
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 18. }\text{If }\tan A+\cot A=4,\text{ then }\tan^4A+\cot^4A\text{ is equal to}
\displaystyle \text{(a) }110\qquad \text{(b) }191\qquad \text{(c) }80\qquad \text{(d) }194
\displaystyle \text{Answer:}
\displaystyle \tan A+\cot A=4
\displaystyle \tan^2A+\cot^2A+2=16
\displaystyle \tan^2A+\cot^2A=14
\displaystyle \tan^4A+\cot^4A+2=196
\displaystyle \tan^4A+\cot^4A=194
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 19. }\text{If }x\sin45^\circ\cos^260^\circ=\frac{\tan^260^\circ\mathrm{cosec}\ 30^\circ}{\sec45^\circ\cot^230^\circ},\text{ then }x=
\displaystyle \text{(a) }2\qquad \text{(b) }4\qquad \text{(c) }8\qquad \text{(d) }16
\displaystyle \text{Answer:}
\displaystyle x\cdot\frac{1}{\sqrt2}\cdot\frac{1}{4}=\frac{3\cdot2}{\sqrt2\cdot3}
\displaystyle \frac{x}{4\sqrt2}=\sqrt2
\displaystyle x=8
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 20. }\text{If }\mathrm{cosec}\ \theta-\cot\theta=\frac{1}{2},\ 0<\theta<\frac{\pi}{2},\text{ then }\cos\theta\text{ is equal to}
\displaystyle \text{(a) }-\frac{3}{5}\qquad \text{(b) }-\frac{5}{3}\qquad \text{(c) }\frac{5}{3}\qquad \text{(d) }\frac{3}{5}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\ \theta-\cot\theta=\frac{1}{2}
\displaystyle \mathrm{cosec}\ \theta+\cot\theta=2
\displaystyle 2\mathrm{cosec}\ \theta=\frac{5}{2}
\displaystyle \mathrm{cosec}\ \theta=\frac{5}{4}
\displaystyle \sin\theta=\frac{4}{5}
\displaystyle \cos\theta=\frac{3}{5}
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 21. }\text{If }\mathrm{cosec}\ A+\cot A=\frac{11}{2},\text{ then }\tan A=
\displaystyle \text{(a) }\frac{21}{22}\qquad \text{(b) }\frac{15}{16}\qquad \text{(c) }\frac{44}{117}\qquad \text{(d) }\frac{117}{43}
\displaystyle \text{Answer:}
\displaystyle \mathrm{cosec}\ A+\cot A=\frac{11}{2}
\displaystyle \mathrm{cosec}\ A-\cot A=\frac{2}{11}
\displaystyle 2\cot A=\frac{11}{2}-\frac{2}{11}=\frac{117}{22}
\displaystyle \cot A=\frac{117}{44}
\displaystyle \therefore \tan A=\frac{44}{117}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 22. }\text{If }\tan\theta+\sec\theta=e^x,\text{ then }\cos\theta\text{ equals}
\displaystyle \text{(a) }\frac{e^x+e^{-x}}{2}\qquad \text{(b) }\frac{2}{e^x+e^{-x}}\qquad \text{(c) }\frac{e^x-e^{-x}}{2}\qquad \text{(d) }\frac{e^x-e^{-x}}{e^x+e^{-x}}
\displaystyle \text{Answer:}
\displaystyle \tan\theta+\sec\theta=e^x
\displaystyle \sec\theta-\tan\theta=e^{-x}
\displaystyle 2\sec\theta=e^x+e^{-x}
\displaystyle \sec\theta=\frac{e^x+e^{-x}}{2}
\displaystyle \therefore \cos\theta=\frac{2}{e^x+e^{-x}}
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 23. }\text{If }\sec\theta+\tan\theta=k,\text{ then }\cos\theta=
\displaystyle \text{(a) }\frac{k^2+1}{2k}\qquad \text{(b) }\frac{2k}{k^2+1}\qquad \text{(c) }\frac{k}{k^2+1}\qquad \text{(d) }\frac{k}{k^2-1}
\displaystyle \text{Answer:}
\displaystyle \sec\theta+\tan\theta=k
\displaystyle \sec\theta-\tan\theta=\frac{1}{k}
\displaystyle 2\sec\theta=k+\frac{1}{k}=\frac{k^2+1}{k}
\displaystyle \sec\theta=\frac{k^2+1}{2k}
\displaystyle \therefore \cos\theta=\frac{2k}{k^2+1}
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 24. }\text{If }f(x)=\cos^2x+\sec^2x,\text{ then}
\displaystyle \text{(a) }f(x)<1\qquad \text{(b) }f(x)=1\qquad \text{(c) }2<f(x)<1\qquad \text{(d) }f(x)\geq2
\displaystyle \text{Answer:}
\displaystyle f(x)=\cos^2x+\frac{1}{\cos^2x}
\displaystyle a+\frac{1}{a}\geq2,\text{ where }a=\cos^2x>0
\displaystyle \therefore f(x)\geq2
\displaystyle \therefore \text{Correct option is (d).}
\\

\displaystyle \textbf{Question 25. }\text{Which of the following is incorrect?}
\displaystyle \text{(a) }\sin\theta=-\frac{1}{5}\qquad \text{(b) }\cos\theta=1\qquad \text{(c) }\sec\theta=\frac{1}{2}\qquad \text{(d) }\tan\theta=20
\displaystyle \text{Answer:}
\displaystyle \sec\theta=\frac{1}{\cos\theta}
\displaystyle -1\leq\cos\theta\leq1
\displaystyle \therefore |\sec\theta|\geq1
\displaystyle \text{So, }\sec\theta=\frac{1}{2}\text{ is not possible}
\displaystyle \therefore \text{Correct option is (c).}
\\

\displaystyle \textbf{Question 26. }\text{The value of }\cos1^\circ\cos2^\circ\cos3^\circ\cdots\cos179^\circ\text{ is}
\displaystyle \text{(a) }\frac{1}{\sqrt2}\qquad \text{(b) }0\qquad \text{(c) }1\qquad \text{(d) }-1
\displaystyle \text{Answer:}
\displaystyle \cos90^\circ=0
\displaystyle \therefore \cos1^\circ\cos2^\circ\cos3^\circ\cdots\cos179^\circ=0
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 27. }\text{The value of }\tan1^\circ\tan2^\circ\tan3^\circ\cdots\tan89^\circ\text{ is}
\displaystyle \text{(a) }0\qquad \text{(b) }1\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) not defined}
\displaystyle \text{Answer:}
\displaystyle \tan\theta\cdot\tan(90^\circ-\theta)=1
\displaystyle \text{All terms pair up, and }\tan45^\circ=1
\displaystyle \therefore \tan1^\circ\tan2^\circ\cdots\tan89^\circ=1
\displaystyle \therefore \text{Correct option is (b).}
\\

\displaystyle \textbf{Question 28. }\text{Which of the following is correct?}
\displaystyle \text{(a) }\sin1^\circ>\sin1\qquad \text{(b) }\sin1^\circ<\sin1\qquad \text{(c) }\sin1^\circ=\sin1\qquad \text{(d) }\sin1^\circ=\frac{\pi}{180}\sin1
\displaystyle \text{Answer:}
\displaystyle 1^\circ=\frac{\pi}{180}\text{ radians}
\displaystyle 0<\frac{\pi}{180}<1
\displaystyle \text{Since }\sin x\text{ is increasing in }(0,1),\ \sin1^\circ<\sin1
\displaystyle \therefore \text{Correct option is (b).}
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\displaystyle \textbf{Question 29. }\text{If }A\text{ lies in second quadrant and }3\tan A+4=0,\text{ then the value of }2\cot A-5\cos A+\sin A\text{ is equal to}
\displaystyle \text{(a) }-\frac{53}{10}\qquad \text{(b) }\frac{23}{10}\qquad \text{(c) }\frac{37}{10}\qquad \text{(d) }\frac{7}{10}
\displaystyle \text{Answer:}
\displaystyle 3\tan A+4=0
\displaystyle \tan A=-\frac{4}{3}
\displaystyle A\text{ lies in second quadrant}
\displaystyle \therefore \sin A=\frac{4}{5},\quad \cos A=-\frac{3}{5},\quad \cot A=-\frac{3}{4}
\displaystyle 2\cot A-5\cos A+\sin A=2\left(-\frac{3}{4}\right)-5\left(-\frac{3}{5}\right)+\frac{4}{5}
\displaystyle =-\frac{3}{2}+3+\frac{4}{5}=\frac{23}{10}
\displaystyle \therefore \text{Correct option is (b).}
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