Inverse Trigonometric Functions Question 63
Question: What is $ sin [cot^{-1}{cos(tan^{-1}}x)] $ where $ x>0 $ , equal to?
Options:
A) $ \sqrt{\frac{(x^{2}+1)}{(x^{2}+2)}} $
B) $ \sqrt{\frac{(x^{2}+2)}{(x^{2}+1)}} $
C) $ \frac{(x^{2}+1)}{(x^{2}+2)} $
D) $ \frac{(x^{2}+2)}{(x^{2}+1)} $
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Answer:
Correct Answer: A
Let $ \alpha ={{\tan }^{-1}}x\Rightarrow \tan \alpha =x $ then $ \cos \alpha =\frac{1}{\sqrt{1+{{\tan }^{2}}\alpha }}=\frac{1}{\sqrt{1+x^{2}}} $
$ \Rightarrow \cos (tan^{-1}x)={ \frac{1}{\sqrt{1+x^{2}}} } $
So, $ {{\cot }^{-1}}\cos (ta{n^{-1}}x)=co{t^{-1}}{ \frac{1}{\sqrt{1+x^{2}}} } $ Let $ {{\cot }^{-1}}( \frac{1}{\sqrt{1+x^{2}}} )=\beta $
$ \Rightarrow \cot \beta =\frac{1}{\sqrt{1+x^{2}}} $ and $ \sin \beta =\frac{1}{\sqrt{1+{{\cot }^{2}}\beta }} $
$ =\frac{\sqrt{1+x^{2}}}{\sqrt{x^{2}+1+1}}=\sqrt{\frac{x^{2}+1}{x^{2}+2}} $
$ \Rightarrow \sin [co{t^{-1}}{cos(ta{n^{-1}})}]=\sqrt{\frac{x^{2}+1}{x^{2}+2}} $
BETA
