Inverse Trigonometric Functions Question 134
Question: $ {{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{n^{-1}}(\sqrt{\cos \alpha })=x $ , then sin x is equal to
Options:
A) $ {{\tan }^{2}}\frac{\alpha }{2} $
B) $ {{\cot }^{2}}\frac{\alpha }{2} $
C) $ \tan \alpha $
D) $ \cot \frac{\alpha }{2} $
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Answer:
Correct Answer: A
$ {{\cot }^{-1}}(\sqrt{\cos \alpha })-ta{n^{-1}}(\sqrt{\cos \alpha })=x $
$ \Rightarrow {{\tan }^{-1}}( \frac{1}{\sqrt{\cos \alpha }} )-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x $
$ \Rightarrow {{\tan }^{-1}}\frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}\sqrt{\cos \alpha }}=x $
$ \Rightarrow {{\tan }^{-1}}\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=x $
$ \Rightarrow \tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }} $
$ \Rightarrow \cot x=\frac{2\sqrt{\cos \alpha }}{1-\cos \alpha } $
$ \Rightarrow cosecx=\frac{1+\cos \alpha }{1-\cos \alpha } $
$ \Rightarrow \sin x=\frac{1-\cos \alpha }{1+\cos \alpha } $
$ \Rightarrow \sin x={{\tan }^{2}}\frac{\alpha }{2} $
BETA
