$\int\limits_{π/6}^{π/3}\frac{\tan x}{\tan x + \cot x}dx$ is equal to

$\frac{\pi}{12}$

The correct answer is Option (4) → $\frac{\pi}{12}$

$\int_{\pi/6}^{\pi/3}\frac{\tan x}{\tan x+\cot x}\,dx$

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

$\frac{\tan x}{\tan x+\cot x} =\tan x\cdot\sin x\cos x =\sin^2 x$

$\int_{\pi/6}^{\pi/3}\sin^2 x\,dx =\int_{\pi/6}^{\pi/3}\frac{1-\cos2x}{2}\,dx$

$=\left[\frac{x}{2}-\frac{\sin2x}{4}\right]_{\pi/6}^{\pi/3}$

$=\left(\frac{\pi}{6}-\frac{\sin\frac{2\pi}{3}}{4}\right) -\left(\frac{\pi}{12}-\frac{\sin\frac{\pi}{3}}{4}\right)$

$=\frac{\pi}{12}$

The value of the integral is $\frac{\pi}{12}$.