Inverse Trigonometric Functions Question 178
Question: $ \sin [{{\cot }^{-1}}(\cos {{\tan }^{-1}}x)] $ =
Options:
A) $ \frac{x}{\sqrt{x^{2}+2}} $
B) $ \frac{x}{\sqrt{x^{2}+1}} $
C) $ \frac{1}{\sqrt{x^{2}+2}} $
D) $ \sqrt{\frac{x^{2}+1}{x^{2}+2}} $
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Answer:
Correct Answer: D
Solution:
$ \sin [{{\cot }^{-1}}(\cos {{\tan }^{-1}}x)] $
$ =\sin [ {{\cot }^{-1}}( \cos {{\cos }^{-1}}\frac{1}{\sqrt{1+x^{2}}} ) ] $
$ =\sin [ {{\cot }^{-1}}\frac{1}{\sqrt{1+x^{2}}} ]=\sin [ {{\sin }^{-1}}\sqrt{\frac{1+x^{2}}{2+x^{2}}} ] $
$ =\sqrt{\frac{1+x^{2}}{2+x^{2}}} $ .
BETA
