d dx [ ^ - 1 ( x) + ^ - 1 ( x)] = - Vidyadip

MCQ

${d \over {dx}}[{\tan ^{ - 1}}(\cot x) + {\cot ^{ - 1}}(\tan x)] = $

A
$0$
B
$1$
C
$-1$
✓
$-2$

✓

Answer

Correct option: D.

$-2$

d
(d) $\frac{d}{{dx}}[{\tan ^{ - 1}}(\cot x) + {\cot ^{ - 1}}(\tan x)]$

=$\frac{{1( - {\rm{cose}}{{\rm{c}}^2}x)}}{{1 + {{\cot }^2}x}} - \frac{{1({{\sec }^2}x)}}{{1 + {{\tan }^2}x}} = - 1 - 1 = - 2$.

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