d dx 1 - 2x 1 + 2x = - Vidyadip

MCQ

${d \over {dx}}\sqrt {{{1 - \sin 2x} \over {1 + \sin 2x}}} = $

A
${\sec ^2}x$
✓
$ - {\sec ^2}\left( {{\pi \over 4} - x} \right)$
C
${\sec ^2}\left( {{\pi \over 4} + x} \right)$
D
${\sec ^2}\left( {{\pi \over 4} - x} \right)$

✓

Answer

Correct option: B.

$ - {\sec ^2}\left( {{\pi \over 4} - x} \right)$

(b) $y = \sqrt {\frac{{1 - \sin 2x}}{{1 + \sin 2x}}} = \frac{{\cos x - \sin x}}{{\cos x + \sin x}}$

$ = \frac{{1 - \tan x}}{{1 + \tan x}} = \tan \left( {\frac{\pi }{4} - x} \right) $

$\Rightarrow \frac{{dy}}{{dx}} = - {\sec ^2}\left( {\frac{\pi }{4} - x} \right)$.

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