Inverse Trigonometric Functions Question 17
Question: If $ u={{\cot }^{-1}}\sqrt{\tan \alpha }-{{\tan }^{-1}}\sqrt{\tan \alpha }, $ then $ \tan ( \frac{\pi }{4}-\frac{u}{2} ) $ is equal to
Options:
A) $ \sqrt{\tan \alpha } $
B) $ \sqrt{\cot \alpha } $
C) $ \tan \alpha $
D) $ \cot \alpha $
Show Answer
Answer:
Correct Answer: A
Let $ \sqrt{\tan \alpha }=\tan x, $ Then $ u={{\cot }^{-1}}(tanx)-ta{n^{-1}}(tanx) $
$ =\frac{\pi }{2}-x-x=\frac{\pi }{2}-2x $
$ \Rightarrow 2x=\frac{\pi }{2}-u\Rightarrow x=\frac{\pi }{4}-\frac{u}{2} $
$ \Rightarrow \tan x=\tan ( \frac{\pi }{4}-\frac{u}{2} )\Rightarrow \sqrt{\tan \alpha }=\tan ( \frac{\pi }{4}-\frac{u}{2} ) $
BETA
