Trigonometric Identities Question 306
Question: If $ \tan x=\frac{b}{a}, $ then $ \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}= $
[MP PET 1990, 2002]
Options:
A) $ \frac{2\sin x}{\sqrt{\sin 2x}} $
B) $ \frac{2\cos x}{\sqrt{\cos 2x}} $
C) $ \frac{2\cos x}{\sqrt{\sin 2x}} $
D) $ \frac{2\sin x}{\sqrt{\cos 2x}} $
Show Answer
Answer:
Correct Answer: B
Solution:
Given that, $ \tan x=\frac{b}{a} $ Now $ \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\sqrt{\frac{1+b/a}{1-b/a}}+\sqrt{\frac{1-b/a}{1+b/a}} $ $ =\frac{2}{\sqrt{1-\frac{b^{2}}{a^{2}}}}=\frac{2}{\sqrt{1-{{\tan }^{2}}x}}=\frac{2}{\sqrt{1-\frac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}}=\frac{2\cos x}{\sqrt{\cos 2x}} $ .
BETA
