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.

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