\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ = ?

Question

Question

This question was previously asked in

SSC CGL 2020 Tier-I Official Paper 5 (Held On : 16 Aug 2021 Shift 2)

Answer 1

Given:

\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ

Formula:

(a + b)² = a² + b² + 2ab

sin²θ + cos²θ = 1

Calculation:

\(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ

⇒ \(cos^3θ\over sin^3θ\) × sin²θ + \(sin^3θ\over cos^3θ\) × cos²θ + 2sinθ cosθ

⇒ \(cos^3θ\over sinθ\) + \(sin^3θ\over cosθ\) + 2sinθ cosθ

⇒ \({sin^4θ+cos^4θ+2sin^2θ cos^2θ}\over{sinθ cosθ}\)

⇒ \(({sin^2θ+cos^2θ})^2\over{sinθ cosθ}\) = \(1\over{sinθ cosθ}\)

⇒ \({1\over sinθ}\times{1\over cosθ}\) = cosecθ secθ (\(1\over{sinθ}\) = cosecθ, \(1\over{cosθ}\) = secθ)

∴ \(\frac{cot^3θ}{cosec^2θ}+\frac{tan^3θ}{sec^2 θ}\) + 2sinθ cosθ = cosecθ.secθ