Question

If $\frac{1+\tan \theta }{1-\tan \theta }=\sqrt{3}$, then find the value of $\theta .$

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${45}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is A ${15}^{\circ }$

Given, $\frac{1+\tan \theta }{1-\tan \theta }=\sqrt{3}$
$\Rightarrow \frac{\tan \frac{\pi }{4}+\tan \theta }{1-\tan \frac{\pi }{4}\tan \theta }=\sqrt{3}$
$\Rightarrow \tan [\frac{\pi }{4}+\theta ]=\sqrt{3}$
$\Rightarrow \frac{\pi }{4}+\theta ={\tan }^{-1}\left[\sqrt{3}\right]$
$\Rightarrow \frac{\pi }{4}+\theta =\frac{\pi }{3}$

$\Rightarrow \theta =\frac{\pi }{3}-\frac{\pi }{4}$

$\Rightarrow \theta =\frac{\pi }{12}$
Therefore, $\theta ={15}^0$