If (a^2 - ­b^2) sin θ + 2 ab cosθ = a^2 + b^2, then the value of tanθ is

Cos(a - b) = cosa cosb + sina sinb

(a$$^2$$ - b$$^2$$) sinθ + 2ab cosθ = a$$^2$$ + b$$^2$$
$$\frac{a^2 - b^2}{a^2 + b^2} sin \theta + \frac{2ab}{a^2 + b^2} cos \theta = a^2 + b^2$$
Let $$\frac{(a^2 - b^2)}{(a^2 - b^2)}$$ = sin A,
then $$ \frac{2ab}{a^2 + b^2} $$= cosA
sinAsinθ + cosAcosθ = 1
cos (A - θ) = 1
A - θ = 0° (as, cos 0° = 1)
θ = A
∴ tanθ = tanA
tanθ = sinA/cosA
tanθ= $$\frac{\frac{(a^2 - b^2)}{(a^2 - b^2)}}{\frac{2ab}{a^2 + b^2}}$$
tanθ= $$\frac{(a^2 - b^2)}{2ab}$$
Option C is the correct answer.