Question

The value of $\frac{(\cos {11}^0+\sin {11}^{\circ })}{(\cos {11}^{\circ }-\sin {11}^0)}$ is

• A. $-\tan {304}^{\circ }$
• B. $\tan {56}^{\circ }$
• C. $\cot {214}^{\circ }$
• D. $\cot {34}^{\circ }$

Answer 1

Answer: (B)

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

Given, $\frac{\cos {11}^o+\sin {11}^o}{\cos {11}^o-\sin {11}^o}$

Divide by $\cos {11}^o$

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

$[\because 1=\tan ({45}^o\left)\right]$

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

$=\tan (45+11)$

$=\tan \left({56}^o\right)$