If (a² - b²) sin θ + 2ab cos θ = a² + b², then find value of cos θ + cot θ

\(\frac{4a^3b}{a^4 - b^4}\)

(By property : sin²θ + cos²θ = 1, while a sin θ + b cos θ = c and a² + b² = c², then sin θ = \(\frac{a}{c}\) and cos θ = \(\frac{b}{c}\)

(a² - b²) sin θ + 2ab cos θ = a² + b²

where, (a² - b²)² + (2ab)² = (a² + b²)²  , therefore

 ⇒ sin θ = \(\frac{(a^2 - b^2)}{a^2 + b^2}\),    cos θ = \(\frac{2ab}{a^2 + b^2}\)

cos θ + cot θ = cosθ + \(\frac{cosθ}{sinθ}\)

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

                    = \(\frac{2a^3b + 2a^3b}{a^4 - b^4}\)

                    = \(\frac{4a^3b}{a^4 - b^4}\)