# Aptitude Question ID : 94434

If $\frac{\tan +\cot }{\tan -\cot } = 2, (0\leq \theta\leq 90^{\circ}),$ then the value of $\sin \theta$ is :
[A]$1$
[B]$\frac{1}{2}$
[C]$\frac{\sqrt{3}}{2}$
[D]$\frac{2}{\sqrt{3}}$

$\mathbf{\frac{\sqrt{3}}{2}}$
$\frac{\tan \theta + \cot \theta}{\tan \theta - \cot \theta} = \frac{2}{1}$
By componendo and dividendo,
$\frac{2\tan \theta}{2\cot \theta} = \frac{3}{1}$
$=> \frac{\sin \theta }{\cos \theta }\cdot \frac{\sin \theta }{\cos \theta } = 3$
$=> \sin ^{2}\theta = 3\cos ^{2}\theta$
$=> \sin ^{2}\theta = 3 \left ( 1-\sin ^{2}\theta \right )$
$=> 4\sin ^{2}\theta = 3$
$=> \sin ^{2}\theta = \frac{3}{4}$
$=> \sin \theta = \frac{\sqrt{3}}{2}$
Hence option [C] is correct answer.