If (a2 - b2) sin θ + 2ab cos θ = a2 + b2, then find value of cos θ + cot θ
Answer & explanation
Correct answer: option 1
(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 + \(\frac{cosθ}{sinθ}\)
=\(\frac{2ab}{a^2 + b^2}\) + \(\frac{2ab}{a^2 - b^2}\)
= \(\frac{2a^3b + 2a^3b}{a^4 - b^4}\)
= \(\frac{4a^3b}{a^4 - b^4}\)