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  4. If (1+ tan θ/1- tan θ)=√3, then find the value of θ

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

Trigonometry

Solution:

$\frac{1+\tan \theta}{1-\tan \theta}=\sqrt{3} $
$ \frac{\tan 45+\tan \theta}{1-\tan 45-\tan \theta}=\sqrt{3} $
$ \tan \left(45^{\circ}+\theta\right)=\tan 60^{\circ}$
$\Rightarrow 45^{\circ}+\theta=60^{\circ} $
$\theta=60^{\circ}-45^{\circ}=15^{\circ}$