Question

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

Answer 1

$LHS=\frac{\cos {20}^{\circ }+\sin {20}^{\circ }}{\cos {20}^{\circ }-\sin {20}^{\circ }}\text{Dividing numerator and denominator by}\cos {20}^{\circ },=\frac{1+\tan {20}^{\circ }}{1-\tan {20}^{\circ }}=\frac{\tan {45}^{\circ }+\tan {20}^{\circ }}{1-\tan {45}^{\circ }\tan {20}^{\circ }}\left[\text{Using identity}\tan \right(\theta +{45}^{\circ })=\frac{1+\tan \theta }{1-\tan \theta }]=\tan ({45}^{\circ }+{20}^{\circ })=\tan {65}^{\circ }=RHS$