Question

Prove that

$$\frac{\cos {11}^{\circ }+\sin {11}^{\circ }}{\cos {11}^{\circ }-\sin {11}^{\circ }}=\tan {56}^{\circ }$$

Answer 1

$$\text{ LHS }=\frac{\cos {11}^{\circ }+\sin {11}^{\circ }}{\cos {11}^{\circ }-\sin {11}^{\circ }}$$
$$=\frac{\frac{\cos {11}^{\circ }}{\cos {11}^{\circ }}+\frac{\sin {11}^{\circ }}{\cos {11}^{\circ }}}{\frac{\cos {11}^{\circ }}{\cos {11}^{\circ }}-\frac{\sin {11}^{\circ }}{\cos {11}^{\circ }}}(\text{ Dividing numerator and denominator by }\cos {11}^{\circ })$$
$$=\frac{1+\tan {11}^{\circ }}{1-\tan {11}^{\circ }}$$
$$=\frac{1+\tan {11}^{\circ }}{1-1\times \tan {11}^{\circ }}$$
$$=\frac{\tan {45}^{\circ }+\tan {11}^{\circ }}{1-\tan {45}^{\circ }\tan {11}^{\circ }}(\text{ As }\tan {45}^{\circ }=1)$$
$$=\tan ({45}^{\circ }+{11}^{\circ })[\text{ As }\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan (A+B)]$$
$$=\tan {56}^{\circ }$$
 = RHS
Hence proved .