Question

Match the terms of Column I with the terms of Column II and choose the correct option from the codes given below.

Column I		Column II
A	$\sin \frac{25\pi }{3}$	1	$-\sqrt{3}$
B	$\cos \frac{41\pi }{4}$	2	$\frac{\sqrt{3}}{2}$
C	$\tan \left(\frac{-16\pi }{3}\right)$	3	$1$
D	$\cot \frac{29\pi }{4}$	4	$\frac{1}{\sqrt{2}}$

• A. A - (2), B -(4), C - (1), D - (3)
• B. A - (2), B -(1), C - (4), D - (3)
• C. A - (3), B -(1), C - (4), D - (2)
• D. A - (3), B -(4), C - (1), D - (2)

Answer 1

Answer: (A)

A. $\sin \frac{25\pi }{3}=\sin \left(8\pi +\frac{\pi }{3}\right)=\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$
$[\because \sin (2n\pi +\theta )=\sin \theta ]$
B. $\cos \frac{41\pi }{4}=\cos \left(10\pi +\frac{\pi }{4}\right)=\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$[\because \cos (2n\pi +\theta )=\cos \theta ]$
C. $\tan \left(\frac{-16\pi }{3}\right)=-\tan \frac{16\pi }{3}[\because \tan (-\theta )=-\tan \theta ]$
$=-\tan \left(5\pi +\frac{\pi }{3}\right)=-\tan \frac{\pi }{3}=-\sqrt{3}$
$[\because \tan (n\pi +\theta )=\tan \theta ]$
D. $\cot \frac{29\pi }{4}=\cot \left(7\pi +\frac{\pi }{4}\right)=\cot \frac{\pi }{4}=1$
$[\because \cot (n\pi +\theta )=\cot \theta ]$
$\therefore A\leftrightarrow 2,B\leftrightarrow 4,C\leftrightarrow 1,D\leftrightarrow 3$