Question

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

Answer 1

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

$=\frac{\cos {11}^{\circ }(1+\tan {11}^{\circ })}{\cos {11}^{\circ }(1-\tan {11}^{\circ })}$ [Taking $\cos {11}^{\circ }$ common ]

$=\frac{\tan {45}^{\circ }+\tan {11}^{\circ }}{1-\tan {45}^{\circ }\tan {11}^{\circ }}$ as $(\tan {45}^{\circ }=1)$

$=\tan ({45}^{\circ }+{11}^{\circ })$ [$\because \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$]

$=\tan {56}^{\circ }=$ RHS

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