A) 0
B) 2
C) 1
D) \[-1\]
E) \[2\sqrt{2}\]
Correct Answer: D
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\[{{\tan }^{-1}}x=-\frac{\pi }{4}\Rightarrow x=-1\]
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