Inverse Trigonometric Functions Question 127
Question: If $ {{\cot }^{-1}}(\sqrt{\cos \alpha )}-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x $ , then sin x is
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 $
Or $ {{\tan }^{-1}}( \frac{1}{\sqrt{\cos \alpha }} )-{{\tan }^{-1}}(\sqrt{\cos \alpha })=x $
Or $ {{\tan }^{-1}}\frac{\frac{1}{\sqrt{\cos \alpha }}-\sqrt{\cos \alpha }}{1+\frac{1}{\sqrt{\cos \alpha }}\sqrt{\cos \alpha }}=x $
Or $ {{\tan }^{-1}}\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=x $
Or $ \tan x=\frac{1-\cos \alpha }{2\sqrt{\cos \alpha }}=x $
Or $ \cot x=\frac{2\sqrt{\cos \alpha }}{1-\cos \alpha } $
Or $ cosecx=\sqrt{1+\frac{4\cos \alpha }{{{(1-cos\alpha )}^{2}}}}=\frac{1+\cos \alpha }{1-\cos \alpha } $
Or $ \sin x=\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{2{{\sin }^{2}}(\alpha /2)}{2{{\cos }^{2}}(\alpha /2)}={{\tan }^{2}}\frac{\alpha }{2} $
BETA
