Question

If b $\tan \theta =a$ find the value of $\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$

• A. $\frac{b+a}{b-a}$
• B. $\frac{b-a}{b+a}$
• C. $\frac{b+a}{a-b}$
• D. None of these

Answer 1

Answer: (A)

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

$b\tan \theta =a\Rightarrow \tan \theta =\frac{a}{b}$
$\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$

$=\frac{\frac{\cos \theta }{\cos \theta }+\frac{\sin \theta }{\cos \theta }}{\frac{\cos \theta }{\cos \theta }-\frac{\sin \theta }{\cos \theta }}$ (Dividing both numerator and denominator by $\cos \theta$)

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

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

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

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