Question

$\underset{n\to \infty }{lim}(\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\frac{1}{2^3}\tan \frac{x}{2^3}+......+\frac{1}{2^n}\tan \frac{x}{2^n})$ is equal to.

• A. $\frac{1}{x}-\tan x$
• B. $\frac{\tan x}{x}$
• C. $\frac{\tan x-x}{x\tan x}$
• D. $x-\cot x$

Answer 1

The correct option is C $\frac{\tan x-x}{x\tan x}$

$\tan 2x=\frac{\sin 2x}{\cos 2x}\Rightarrow \frac{1}{\cot 2x}=\frac{2\sin x\cos x}{{\cos }^{2}x-{\sin }^{2}x}$

$\Rightarrow \cot 2x=\frac{{\cos }^{2}x-{\sin }^{2}x}{2\sin x\cos x}\Rightarrow 2\cot 2x=\frac{{\cos }^{2}x}{\sin x\cos x}-\frac{{\sin }^{2}x}{\sin x\cos x}$

$\Rightarrow 2\cot 2x=\cot x-\tan x\Rightarrow \tan x=\cot x-2\cot 2x$

$\Rightarrow \frac{1}{2}\tan \frac{x}{2}=\frac{1}{2}\cot \frac{x}{2}-\cot x$$\Rightarrow \frac{1}{2^2}\tan \frac{x}{2^2}=\frac{1}{2^2}\cot \frac{x}{2}-\frac{1}{2}\cot x$

Similarly $\Rightarrow \frac{1}{2^n}\tan \frac{x}{2^n}=\frac{1}{2^n}\cot \frac{x}{2^n}-\frac{1}{2^{n-1}}\cos \frac{x}{2^{n-1}}$

Let $S=(\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}+\frac{1}{2^3}\tan \frac{x}{2^3}+...+\frac{1}{2^n}\tan \frac{x}{2^n})$

$\Rightarrow \underset{n\to \infty }{lim}S=\underset{n\to \infty }{lim}(\frac{1}{2}\cot \frac{x}{2}-\cot x+\frac{1}{2^n}\cot \frac{x}{2^n}-\frac{1}{2^{n-1}})$

$=\underset{n\to \infty }{lim}(-\cot x+\frac{1}{2^n}\cot \frac{x}{2^n})=\frac{\tan x-x}{x\tan x}$