If tan θ=ab, then a sin θ+b cos θa sin θ−b cos θis equal to

Question:

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

Solution:

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 $(a)$