d dx [ ( ^2 2x - ^2 x 1 - ^2 2x ^2 x ) 3x ] = - Vidyadip

MCQ

${d \over {dx}}\left[ {\left( {{{{{\tan }^2}2x - {{\tan }^2}x} \over {1 - {{\tan }^2}2x{{\tan }^2}x}}} \right)\cot 3x} \right] =$

A
$\tan 2x\,\tan x$
B
$\tan 3x\tan x$
✓
${\sec ^2}x$
D
$\sec x\tan x$

✓

Answer

Correct option: C.

${\sec ^2}x$

c
(c) Let $y = \frac{{{{\tan }^2}2x - {{\tan }^2}x}}{{1 - {{\tan }^2}2x{{\tan }^2}x}}$

$= \frac{{(\tan 2x - \tan x)}}{{(1 + \tan 2x\tan x)}}\,\frac{{(\tan 2x + \tan x)}}{{(1 - \tan 2x\tan x)}}$

$= \tan (2x - x)\,\tan (2x + x)$$ = \tan x\tan 3x$.

$\therefore$ $\frac{d}{{dx}}[y.\cot 3x] = \frac{d}{{dx}}[\tan x] = {\sec ^2}x$.

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