Inverse Trigonometric Functions Question 8
Question: $ {{\cot }^{-1}}[{{(\cos \alpha )}^{1/2}}]-{{\tan }^{-1}}[{{(\cos \alpha )}^{1/2}}]=x, $ then $ \sin x= $
[AIEEE 2002]
Options:
A) $ {{\tan }^{2}}( \frac{\alpha }{2} ) $
B) $ {{\cot }^{2}}( \frac{\alpha }{2} ) $
C) $ \tan \alpha $
D) $ \cot ( \frac{\alpha }{2} ) $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{\tan }^{-1}}[ \frac{1}{\sqrt{\cos \alpha }} ]-{{\tan }^{-1}}[ \sqrt{\cos \alpha } ]=x $
Therefore $ {{\tan }^{-1}}[ \frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{\sqrt{\cos \alpha }}{\sqrt{\cos \alpha }}} ]=x $
Therefore $ \tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }} $
$ \therefore \sin x=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{{\sin }^{2}}\frac{\alpha }{2}}{2{{\cos }^{2}}\frac{\alpha }{2}}={{\tan }^{2}}( \frac{\alpha }{2} ) $ .
BETA
