Inverse Trigonometric Functions Question 216
Question: If $ {{\tan }^{-1}}\frac{a+x}{a}+{{\tan }^{-1}}\frac{a-x}{a}=\frac{\pi }{6} $ ,then $ x^{2} $ =
Options:
A) $ 2\sqrt{3}a $
B) $ \sqrt{3}a $
C) $ 2\sqrt{3}a^{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Given equation is $ {{\tan }^{-1}}\frac{a+x}{a}+{{\tan }^{-1}}\frac{a-x}{a}=\frac{\pi }{6} $
$ \Rightarrow {{\tan }^{-1}}( \frac{\frac{a+x}{a}+\frac{a-x}{a}}{1-\frac{a+x}{a}.\frac{a-x}{a}} )=\frac{\pi }{6} $
$ \Rightarrow \frac{2a^{2}}{x^{2}}=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}\Rightarrow x^{2}=2\sqrt{3}a^{2} $ .
BETA
