If $\tan \theta=\frac{a}{b}$, then $\frac{a \sin \theta+b \cos \theta}{a \sin \theta-b \cos \theta}$ is equal to
(a) $\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$
(b) $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$
(C) $\frac{a+b}{a-b}$
(s) $\frac{a-b}{a+b}$
Given: $\tan \theta=\frac{a}{b}$
We have to find the value of following expression in terms of a and b
We know that: $\tan \theta=\frac{\text { Perpendicular }}{\text { Base }}$
$\Rightarrow$ Base $=b$
$\Rightarrow$ Perpendicular $=a$
$\Rightarrow$ Hypotenuse $=\sqrt{(\text { Perpendicular })^{2}+(\text { Base })^{2}}$
$\Rightarrow$ Hypotenuse $=\sqrt{a^{2}+b^{2}}$
Now we find,
$\frac{a \sin \theta+b \cos \theta}{a \sin \theta-b \cos \theta}=\frac{a\left(\frac{a}{a^{2}+b^{2}}\right)+b\left(\frac{b}{a^{2}+b^{2}}\right)}{a\left(\frac{a}{a^{2}+b^{2}}\right)-b\left(\frac{b}{a^{2}+b^{2}}\right)}$
$=\frac{\frac{a^{2}+b^{2}}{a^{2}+b^{2}}}{\frac{a^{2}-b^{2}}{a^{2}+b^{2}}}$
$=\frac{a^{2}+b^{2}}{a^{2}-b^{2}}$
Hence the correct option is $(a)$