If $\tan \theta+\cot \theta=2$, then $\sin \theta$ is equal to
If $\tan \theta+\cot \theta=2$, then $\sin \theta$ is equal to
- $\frac{1}{\sqrt{2}}$
- $\frac{1}{\sqrt{3}}$
- $\frac{1}{2}$
- $1$
Solution
We have,
$\begin{array}{rlrl} & & \tan \theta+\cot \theta & =2 \\ \Rightarrow & & \frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta} & =2 \\ \Rightarrow & \frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta \cos \theta} & =2 \\ \Rightarrow & 1=2 \sin \theta \cos \theta & \Rightarrow \sin 2 \theta=1 \\ \Rightarrow & 2 \theta & =\frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{4}\end{array}$
Thus, $\quad \sin \theta=\sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}$
Asked in: AP EAMCET 2001
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