Inverse Trigonometric Functions Question 197
Question: If $ {{\tan }^{-1}}\frac{x-1}{x+2}+{{\tan }^{-1}}\frac{x+1}{x+2}=\frac{\pi }{4} $ , then x =
Options:
A) $ \frac{1}{\sqrt{2}} $
B) $ -\frac{1}{\sqrt{2}} $
C) $ \pm \sqrt{\frac{5}{2}} $
D) $ \pm \frac{1}{2} $
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Answer:
Correct Answer: C
Solution:
We have $ {{\tan }^{-1}}\frac{x-1}{x+2}+{{\tan }^{-1}}\frac{x+1}{x+2}=\frac{\pi }{4} $
$ \Rightarrow {{\tan }^{-1}}[ \frac{\frac{x-1}{x+2}+\frac{x+1}{x+2}}{1-( \frac{x-1}{x+2} )( \frac{x+1}{x+2} )} ]=\frac{\pi }{4} $
$ \Rightarrow [ \frac{2x(x+2)}{x^{2}+4+4x-x^{2}+1} ]=\tan \frac{\pi }{4} $
Therefore $ \frac{2x(x+2)}{4x+5}=\tan \frac{\pi }{4}=1 $
$ \Rightarrow 2x^{2}+4x=4x+5 $
$ \Rightarrow x=\pm \sqrt{\frac{5}{2}} $ .
BETA
