Question

$\underset{\theta \to 0}{lim}\frac{4\theta (\tan \theta -2\theta \tan \theta )}{(1-\cos 2\theta )}$ is.

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

Answer 1

The correct option is D $2$

$\underset{\theta \to 0}{lim}\frac{4\theta (\tan \theta -2\theta \tan \theta )}{1-\cos 2\theta }$
$=\underset{\theta \to 0}{lim}\frac{4\theta (\tan \theta -2\theta \tan \theta )}{2{\sin }^2\theta }$
$=\underset{\theta \to 0}{lim}\frac{2\theta (\tan \theta -2\theta \tan \theta )}{\frac{{\sin }^2\theta }{{\theta }^2}\times {\theta }^2}$
$=\underset{\theta \to 0}{lim}\frac{2}{{\left[\frac{\sin \theta }{\theta }\right]}^2}\times \left[\frac{\tan \theta -2\theta \tan \theta }{\theta }\right]$

$=\underset{\theta \to 0}{lim}2[\frac{\tan \theta }{\theta }-2\tan \theta ]$ ........$[\because \underset{x\to 0}{lim}\frac{\sin x}{x}]$
$=2[1-2(0\left)\right]$ ........ $[\because \underset{x\to 0}{lim}\frac{\tan x}{x}=1]$
$=2$