Trigonometric Identities Question 196
Question: $ \sqrt{3},cosec,20^{o}-\sec ,20^{o}= $
[IIT 1988]
Options:
A) 2
B) $ \frac{2,\sin 20^{o}}{\sin 40^{o}} $
C) 4
D) $ \frac{4,\sin 20^{o}}{\sin 40^{o}} $
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Answer:
Correct Answer: C
Solution:
$ \sqrt{3}cosec,20{}^\circ -\sec 20{}^\circ =\frac{\sqrt{3}}{\sin 20{}^\circ }-\frac{1}{\cos ,20{}^\circ } $ $ =\frac{\sqrt{3}\cos 20{}^\circ -\sin 20{}^\circ }{\sin 20{}^\circ \cos 20{}^\circ }=\frac{2[ \frac{\sqrt{3}}{2}\cos 20{}^\circ -\frac{1}{2}\sin ,20{}^\circ ]}{\frac{2}{2}\sin 20{}^\circ \cos 20{}^\circ } $ $ =\frac{4\cos (20{}^\circ +30{}^\circ )}{\sin 40{}^\circ }=\frac{4\cos 50{}^\circ }{\sin 40{}^\circ }=\frac{4\sin 40{}^\circ }{\sin 40{}^\circ }=4 $ .
BETA
