Question

${}_{x\to 0}^{lim}\frac{x\tan 2x-2x\tan x}{{(1-\cos 2x)}^2}$ equals $:-$

• A. $-\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{2}$

Given that.

${lim}_{x\to 0}\frac{x\tan 2x-2x\tan x}{{(1-\cos 2x)}^2}$

${lim}_{x\to 0}\frac{x.\frac{2\tan x}{1-{\tan }^{2}x}-2x\tan x}{{(1-1+2{\sin }^{2}x)}^2}\therefore \tan 2\theta =\frac{2tan\theta }{1-{\tan }^2\theta }$

${lim}_{x\to 0}\frac{2x\tan x-2x\tan x+2x{\tan }^{3}x}{(1-{\tan }^{2}x){\left(2{\sin }^{2}x\right)}^2}$

${lim}_{x\to 0}\frac{2x{\tan }^{2}x}{(1-{\tan }^{2}x)4{\sin }^{4}x}$

${lim}_{x\to 0}\frac{x{\tan }^{3}x}{2{\sin }^{4}x}{lim}_{x\to 0}\frac{1}{1-{\tan }^{2}x}$

${lim}_{x\to 0}\frac{\frac{{\tan }^{3}x}{x^3}}{2\frac{{\sin }^{4}x}{x^4}}\times 1$

$=\frac{1}{2\times 1}$

$=\frac{1}{2}$

Hence this is the answer.