If (a2 - b2) sinx + ab cosx = a2 + b2 then find the value of cosθ.tanθ

\(\frac{a^2-b^2}{2ab}\) \(\frac{a^2+b^2}{a^2-b^2}\) \(\frac{a^2-b^2}{a^2+b^2}\) \(\frac{a+b}{a^2-b^2}\)

\(\frac{a^2-b^2}{a^2+b^2}\)

Make sin2θ + cos2θ = 1 form so divide the equation by a2 + b2 \(\frac{a^2-b^2}{a^2+b^2}\)sinx + \(\frac{2ab}{a^2+b^2}\)cosx = 1 P = a2 - b2 H = a2 + b2 B = 2ab ⇒ cosx.tanx = \(\left(\frac{2ab}{a^2+b^2} \right) \left(\frac{a^2-b^2}{2ab} \right)\) = \(\frac{a^2-b^2}{a^2+b^2}\)