if tanθ = a/b then the value of b sinθ - a cosθ / b sin θ + a cos θ is

Question

If tanθ = a/b then the value of \(\cfrac{bsin\theta-acos\theta}{bsin\theta+acos\theta}\) is

(1) 1

(2) \(\frac{a^2-b^2}{a^2+b^2}\)

(3) \(\frac{b^2-a^2}{b^2+a^2}\)

(4) 0

Answer 1

(4) 0

\(\cfrac{bsin\theta-acos\theta}{bsin\theta+acos\theta}\)

Dividing Nr. & Dr. by cosθ

= \(\cfrac{btan\theta-a}{btan\theta+a}=\) \(\cfrac{b\times\frac{a}{b}-a}{b\times\frac{a}{b}+a}\) = \(\cfrac{0}{2a}\) = 0