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

MCQ

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

A
$1$
✓
$1/2$
C
$\cos x$
D
$\sec x$

✓

Answer

Correct option: B.

$1/2$

b
(b) $\frac{d}{{dx}}{\tan ^{ - 1}}(\sec x + \tan x) = \frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{{1 + \sin x}}{{\cos x}}} \right)$

$ = \frac{d}{{dx}}{\tan ^{ - 1}}\left( {\frac{{\sin \left( {\frac{x}{2}} \right) + \cos \left( {\frac{x}{2}} \right)}}{{\cos \left( {\frac{x}{2}} \right) - \sin \left( {\frac{x}{2}} \right)}}} \right) = \frac{d}{{dx}}\left( {\frac{\pi }{4} + \frac{x}{2}} \right) = \frac{1}{2}$.

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