Question

Prove that $\frac{\cos {10}^0+\sin {10}^0}{\cos {10}^0-\sin {10}^0}=\tan a^0$
Find a

Answer 1

$\frac{\cos {10}^0+\sin {10}^0}{\cos {10}^0-\sin {10}^0}=\frac{1+\tan {10}^0}{1-\tan {10}^0}=\frac{\tan {45}^0+\tan {10}^0}{1-\tan {45}^0\tan {10}^0}=\tan \left({45}^0{10}^0\right)=\tan {55}^0$ (dividing by $\cos {10}^0$)
Ans: 55