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  4. If ( tan -1x)2+( cot -1x)2=(5π 2/8), then x is equal to

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Q. If $ {{({{\tan }^{-1}}x)}^{2}}+{{({{\cot }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8}, $ then $ x $ is equal to

KEAMKEAM 2008

Solution:

Given, $ {{({{\tan }^{-1}}x)}^{2}}+{{({{\cot }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8} $
$ \therefore $ $ {{({{\tan }^{-1}}x+{{\cot }^{-1}}x)}^{2}}-2{{\tan }^{-1}}x\left( \frac{\pi }{2}-{{\tan }^{-1}}x \right) $
$=\frac{5{{\pi }^{2}}}{8} $
$ \Rightarrow $ $ \frac{{{\pi }^{2}}}{4}-2\times \frac{\pi }{2}{{\tan }^{-1}}x+2{{({{\tan }^{-1}}x)}^{2}}=\frac{5{{\pi }^{2}}}{8} $
$ \Rightarrow $ $ 2{{({{\tan }^{-1}}x)}^{2}}-\pi {{\tan }^{-1}}x-\frac{3{{\pi }^{2}}}{8}=0 $
$ \Rightarrow $ $ {{\tan }^{-1}}x=-\frac{\pi }{4},\frac{3\pi }{4} $
Now, we take $ {{\tan }^{-1}}x=-\frac{\pi }{4}\Rightarrow x=-1 $