Question

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

Answer 1

Simplify the LHS of $\frac{\cos {11}^{\circ }+\sin {11}^{\circ }}{\cos {11}^{\circ }-\sin {11}^{\circ }}=\tan {56}^{\circ }$ by dividing the numerator and denominator by $\cos {11}^{\circ }$.

$\frac{\frac{\cos {11}^{\circ }}{\cos {11}^{\circ }}+\frac{\sin {11}^{\circ }}{\cos {11}^{\circ }}}{\frac{\cos {11}^{\circ }}{\cos {11}^{\circ }}-\frac{\sin {11}^{\circ }}{\cos {11}^{\circ }}}=\frac{1+\tan {11}^{\circ }}{1-\tan {11}^{\circ }}$

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

$=\tan \left({45}^{\circ }+{11}^{\circ }\right)$

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

$=RHS$