Question

If $\tan \theta =\frac{a}{b}$, then $\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$ is

• A. $\frac{b+a}{b-a}$
• B. $\frac{b-a}{b+a}$
• C. $\frac{a}{b}$
• D. $\frac{b}{a}$

Answer 1

The correct option is A $\frac{b+a}{b-a}$

Given, $\tan \theta =\frac{a}{b}$

Therefore,$\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$

dividing by $\cos \theta$ in numerator and denominator

$=\frac{1+\tan \theta }{1-\tan \theta }$

$=\frac{1+\frac{a}{b}}{1-\frac{a}{b}}$

$=\frac{\frac{b+a}{b}}{\frac{b-a}{b}}$

$=\frac{b+a}{b-a}$