Prove that cot cube theta into sin cube theta upon cos theta + sin theta whole square + 10 cube theta into cos cube theta upon cos theta + sin theta whole square equal to sec theta cosec theta minus one upon cosec theta + sec theta
Answers
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u have to solve both LHS and RHS
(Cot³θ* Sin³θ)/(Cosθ + Sinθ)² + (Tan³θ* Cos³θ)/(Cosθ + Sinθ)² = (SecθCosecθ - 1)/(Cosecθ + Secθ)
Step-by-step explanation:
(Cot³θ* Sin³θ)/(Cosθ + Sinθ)² + (Tan³θ* Cos³θ)/(Cosθ + Sinθ)² = (SecθCosecθ - 1)/(Cosecθ + Secθ)
LHS =
(Cot³θ* Sin³θ)/(Cosθ + Sinθ)² + (Tan³θ* Cos³θ)/(Cosθ + Sinθ)²
= Cos³θ/(Cosθ + Sinθ)² + (Sin³θ)/(Cosθ + Sinθ)²
= (Cos³θ +Sin³θ)/(Cosθ + Sinθ)²
using a³ + b³ = (a + b)(a² + b² - ab)
= (Cosθ + Sinθ)(Cos²θ + Sin²θ - CosθSinθ)/(Cosθ + Sinθ)²
= (1 - CosθSinθ)/(Cosθ + Sinθ)
= (1 - CosθSinθ)/(1/Secθ + 1/Cosecθ)
= SecθCosecθ(1 - CosθSinθ)/(Cosecθ + Secθ)
= (SecθCosecθ - 1)/(Cosecθ + Secθ)
= RHS
(Cot³θ* Sin³θ)/(Cosθ + Sinθ)² + (Tan³θ* Cos³θ)/(Cosθ + Sinθ)² = (SecθCosecθ - 1)/(Cosecθ + Secθ)
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