A) \[\sqrt{\tan \alpha }\]
B) \[\sqrt{\cot \alpha }\]
C) \[\tan \alpha \]
D) \[\cot \alpha \]
Correct Answer: A
Solution :
[a] Let \[\sqrt{\tan \alpha }=\tan x,\] Then \[u={{\cot }^{-1}}(tanx)-ta{{n}^{-1}}(tanx)\] \[=\frac{\pi }{2}-x-x=\frac{\pi }{2}-2x\] \[\Rightarrow 2x=\frac{\pi }{2}-u\Rightarrow x=\frac{\pi }{4}-\frac{u}{2}\] \[\Rightarrow \tan x=\tan \left( \frac{\pi }{4}-\frac{u}{2} \right)\Rightarrow \sqrt{\tan \alpha }=\tan \left( \frac{\pi }{4}-\frac{u}{2} \right)\]
You need to login to perform this action.
You will be redirected in
3 sec
