Trigonometric Identities Question 25
Question: $ \frac{\sin 70{}^\circ +\cos 40{}^\circ }{\cos 70{}^\circ +\sin 40{}^\circ }= $
[CET 1986; MP PET 1999]
Options:
A) 1
B) $ \frac{1}{\sqrt{3}} $
C) $ \sqrt{3} $
D) $ \frac{1}{2} $
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Answer:
Correct Answer: C
Solution:
$ \frac{\sin 70^{o}+\cos 40^{o}}{\cos 70^{o}+\sin 40^{o}} $ $ =\frac{\sin 70{}^\circ +\sin 50{}^\circ }{\sin 20{}^\circ +\sin 40{}^\circ }=\frac{2\sin 60{}^\circ \cos 10{}^\circ }{2\sin 30{}^\circ \cos (-10{}^\circ )} $ $ =\frac{\sin 60^{o}}{\sin 30^{o}}=\frac{\sqrt{3}}{2}.\frac{2}{1}=\sqrt{3} $ .
BETA
