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

MCQ

${d \over {dx}}\left( {{{\sec x + \tan x} \over {\sec x - \tan x}}} \right) = $

✓
${{2\cos x} \over {{{(1 - \sin x)}^2}}}$
B
${{\cos x} \over {{{(1 - \sin x)}^2}}}$
C
${{2\cos x} \over {1 - \sin x}}$
D
એકપણ નહીં

✓

Answer

Correct option: A.

${{2\cos x} \over {{{(1 - \sin x)}^2}}}$

(a) $\frac{d}{{dx}}\left( {\frac{{\sec x + \tan x}}{{\sec x - \tan x}}} \right) = \frac{d}{{dx}}\left( {\frac{{1 + \sin x}}{{1 - \sin x}}} \right)$.

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