If ( ^ - 1 x)^2 + ( ^ - 1 x)^2 = 5 ^2 8 , then x equals - Vidyadip

MCQ

If ${({\tan ^{ - 1}}x)^2} + {({\cot ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8},$ then $x$ equals

✓
$-1$
B
$1$
C
$0$
D
None of these

✓

Answer

Correct option: A.

$-1$

a
(a) ${({\tan ^{ - 1}}x)^2} + {({\cot ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$

==> ${({\tan ^{ - 1}}x + {\cot ^{ - 1}}x)^2} - 2{\tan ^{ - 1}}x\left( {\frac{\pi }{2} - {{\tan }^{ - 1}}x} \right) = \frac{{5{\pi ^2}}}{8}$

==> $\frac{{{\pi ^2}}}{4} - 2 \times \frac{\pi }{2}{\tan ^{ - 1}}x + 2{({\tan ^{ - 1}}x)^2} = \frac{{5{\pi ^2}}}{8}$

==> $2{({\tan ^{ - 1}}x)^2} - \pi {\tan ^{ - 1}}x - \frac{{3{\pi ^2}}}{8} = 0$

==> ${\tan ^{ - 1}}x = - \frac{\pi }{4},\frac{{3\pi }}{4}$

==> ${\tan ^{ - 1}}x = - \frac{\pi }{4} \Rightarrow x = - 1$.

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