MCQ
If ${({\tan ^{ - 1}}x)^2} + {({\cot ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8},$ then $x$ equals
- ✓$-1$
- B$1$
- C$0$
- DNone of these
==> ${({\tan ^{ - 1}}x + {\cot ^{ - 1}}x)^2} - 2{\tan ^{ - 1}}x\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}x} \right) = \frac{{5{\pi ^2}}}{8}$
==> $\frac{{{\pi ^2}}}{4} - 2 \times \frac{\pi }{2}{\tan ^{ - 1}}x + 2{({\tan ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$
==> $2{({\tan ^{ - 1}}x)^2} - \pi {\tan ^{ - 1}}x - \frac{{3{\pi ^2}}}{8} = 0$
==> ${\tan ^{ - 1}}x = - \frac{\pi }{4},\frac{{3\pi }}{4}$
==> ${\tan ^{ - 1}}x = - \frac{\pi }{4} \Rightarrow x = - 1$.
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