Question

Prove
$\tan {55}^o=\frac{\cos {10}^o+\sin {10}^o}{\cos {10}^o-\sin {10}^o}$

Answer 1

$\tan 55°=\cos 10°+\sin 10°$

$\tan (45°+10°)=\frac{\tan 45°+\tan 10°}{1-\tan 45°\tan 10°}$

$=\frac{1+\frac{\sin 10°}{\cos 10°}}{1-\frac{\sin 10°}{\cos 10°}}$

$=\frac{\cos 10°+\sin 10°}{\cos 10°-\sin 10°}$