Question

$\frac{\cos {10}^0+\sin {10}^0}{\cos {10}^0-\sin {10}^0}$ is equal to

• A. $\tan {55}^0$
• B. $\cot {55}^0$
• C. $-\tan {35}^0$
• D. $-\cot {35}^0$

Answer 1

The correct option is A $\tan {55}^0$

$\frac{\cos {10}^o+\sin {10}^o}{\cos {10}^o-\sin {10}^o}=\frac{1+\frac{\sin {10}^o}{\cos {10}^o}}{1-\frac{\sin {10}^o}{\cos {10}^o}}=\frac{1+\tan {10}^o}{1-\tan {10}^o}=\frac{\tan {45}^o+\tan {10}^o}{1-\tan {45}^o\tan {10}^o}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\tan (45+10)$

$=\tan {55}^o$