Inverse Trigonometric Functions Question 145
Question: If $ \tan ({{\cos }^{-1}}x)=\sin ( {{\cot }^{-1}}\frac{1}{2} ) $ , then x =
Options:
A) $ \pm \frac{5}{3} $
B) $ \pm \frac{\sqrt{5}}{3} $
C) $ \pm \frac{5}{\sqrt{3}} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Given that $ \tan {{{\cos }^{-1}}(x)}=\sin ( {{\cot }^{-1}}\frac{1}{2} ) $ Let $ {{\cot }^{-1}}\frac{1}{2}=\varphi \Rightarrow \frac{1}{2}=\cot \varphi $
$ \Rightarrow \sin \varphi =\frac{1}{\sqrt{1+{{\cot }^{2}}\varphi }}=\frac{2}{\sqrt{5}} $ Let $ {{\cos }^{-1}}x=\theta \Rightarrow \sec \theta =\frac{1}{x}\Rightarrow \tan \theta =\sqrt{{{\sec }^{2}}\theta -1} $
$ \Rightarrow \tan \theta =\sqrt{\frac{1}{x^{2}}-1}\Rightarrow \tan \theta =\frac{\sqrt{1-x^{2}}}{x} $ So, $ \tan {{{\cos }^{-1}}(x)}=\sin ( {{\cot }^{-1}}\frac{1}{2} ) $
$ \Rightarrow \tan ( {{\tan }^{-1}}\frac{\sqrt{1-x^{2}}}{x} )=\sin ( {{\sin }^{-1}}\frac{2}{\sqrt{5}} ) $
$ \Rightarrow \frac{\sqrt{1-x^{2}}}{x}=\frac{2}{\sqrt{5}}\Rightarrow \sqrt{(1-x^{2})5}=2x $ Squaring both sides, we get $ x=\pm \frac{\sqrt{5}}{3} $ .
BETA
