Question

Prove that $\frac{\cos {10}^o-\sin {10}^o}{\cos {10}^o+\sin {10}^o}=\tan {35}^o$.

Answer 1

Now,

$\frac{\cos {10}^o-\sin {10}^o}{\cos {10}^o+\sin {10}^o}$

$=\frac{1-\tan {10}^o}{1+\tan {10}^o}$ [ Dividing the numerator and the denominator by $\cos {10}^o$]

$=\frac{\tan {45}^o-\tan {10}^o}{1+\tan {45}^o.\tan {10}^o}$

$=\tan ({45}^o-{10}^o)$

$=\tan {35}^o$.