Question

Show that $\tan 15=2-\sqrt{3}$

Answer 1

We know that,

$\tan (A.B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$\tan \left(15\right)=\tan (60-40)$

$\frac{\sqrt{3}-1}{1+\sqrt{3}.1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$

Rationalize:-

$\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{3+1-2\sqrt{3}}{3-1}=2-\sqrt{3}$

$\tan \left(15\right)=2-\sqrt{3}$