If $$tan\spaceθ=\tfrac{b}{a}$$​, then $$\tfrac{a\space cos \space θ\space+\space b\space sin\space θ}{{a\space cos \space θ\space-\space b\space sin\space θ}}$$​ = __________

Correct option is C

Given:
$$tan\spaceθ=\frac{b}{a}$$​​, then $$\tfrac{a\space cos \space θ\space+\space b\space sin\space θ}{{a\space cos \space θ\space-\space b\space sin\space θ}}$$​​ = _________

Solution:

$$\frac{a\space cos \space θ\space+\space b\space sin\space θ}{{a\space cos \space θ\space-\space b\space sin\space θ}}$$

Divided by cos θ ,we have​
​= $$\frac{a\space \space+\space b\space tan\space θ}{{a\space\space-\space b\space tan\space θ}}$$

=$$\frac{a\space \space+\frac{b^2}{a}}{{a\space\space-\frac{b^2}{a}}}$$

=$$\frac{a^2+b^2}{a^2-b^2}$$

Option (C) is right.