Trigonometric Identities Question 40
Question: The value of $ \cot 70^{o}+4\cos 70^{o} $ is
[Orissa JEE 2003]
Options:
A) $ \frac{1}{\sqrt{3}} $
B) $ \sqrt{3} $
C) $ 2\sqrt{3} $
D) $ \frac{1}{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Now, $ \cot 70^{o}+4\cos 70^{o}=\frac{\cos 70^{o}+4\sin 70^{o}\cos 70^{o}}{\sin 70^{o}} $ $ =\frac{\cos 70^{o}+2\sin 140^{o}}{\sin 70^{o}}=\frac{\cos 70^{o}+2\sin (180^{o}-40^{o})}{\sin 70^{o}} $ $ =\frac{\sin 20^{o}+\sin 40^{o}+\sin 40^{o}}{\sin 70^{o}}=\frac{2\sin 30^{o}\cos 10^{o}+\sin 40^{o}}{\sin 70^{o}} $ $ =\frac{\sin 80^{o}+\sin 40^{o}}{\sin 70^{o}}=\frac{2\sin 60^{o}\cos 20^{o}}{\sin 70^{o}}=\sqrt{3} $ .
BETA
