Question:
If $\tan \theta=\frac{a}{b}$, then $\frac{(\cos \theta+\sin \theta)}{(\cos \theta-\sin \theta)}=?$
(a) $\frac{a+b}{a-b}$
(b) $\frac{a-b}{a+b}$
(c) $\frac{b+a}{b-a}$
(d) $\frac{b-a}{b+a}$
Solution:
(c) $\frac{b+a}{b-a}$
Given: $\tan \theta=\frac{a}{b}$
Now, $\frac{(\cos \theta+\sin \theta)}{(\cos \theta-\sin \theta)}$
$=\frac{(1+\tan \theta)}{(1-\tan \theta)} \quad[$ Dividing the numerator and denominator by $\cos \theta]$
$=\frac{\left(1+\frac{a}{b}\right)}{\left(1-\frac{a}{b}\right)}$
$=\frac{\left(\frac{b+a}{b}\right)}{\left(\frac{b-a}{b}\right)}$
$=\frac{(b+a)}{(b-a)}$
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