Question

Prove that $\frac{\cos {20}^o+\sin {20}^o}{\cos {20}^o-\sin {20}^o}=\tan {65}^o$.

Answer 1

Now,

$\frac{\cos {20}^o+\sin {20}^o}{\cos {20}^o-\sin {20}^o}$

$=\frac{1+\tan {20}^o}{1-\tan {20}^o}$ [ Dividing the numerator and denominator by $\cos {20}^o$]

$=\frac{\tan {45}^o+\tan {20}^o}{1-\tan {45}^o.\tan {20}^o}$

$=\tan ({45}^o+{20}^o)$

$=\tan {65}^o$.