Question : If $\frac{\tan\theta +\cot\theta }{\tan\theta -\cot\theta }=2, (0\leq \theta \leq 90^{0})$, then the value of $\sin\theta$ is:
Option 1: $\frac{2}{\sqrt3}$
Option 2: $\frac{\sqrt3}{2}$
Option 3: $\frac{1}{2}$
Option 4: $1$
Correct Answer: $\frac{\sqrt3}{2}$
Solution :
$\frac{\tan\theta +\cot\theta }{\tan\theta -\cot\theta }=2$
Use rule of componendo and dividendo i.e. If $\frac{a}{b}=\frac{c}{d}$ then, $\frac{a+b}{a-b}=\frac{c+d}{c-d}$. we have,
⇒ $\frac{\tan\theta +\cot\theta +\tan\theta -\cot\theta}{\tan\theta +\cot\theta -\tan\theta +\cot\theta}=\frac{2+1}{2-1}$
⇒$\frac{2\tan\theta}{2\cot\theta} = 3$
⇒ $\tan^2\theta=3$
⇒ $\sin^2\theta= 3\cos^2\theta$
⇒ $\sin^2\theta= 3(1-\sin^2\theta)$
⇒ $4\sin^2\theta= 3$
⇒ $\sin\theta=\frac{ \sqrt3 }{2}$
Hence, the correct answer is $\frac{ \sqrt3 }{2}$.
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