Cot cube theta into sin cube theta divided by cos theta + sin theta whole square + tan cube theta into cos cube theta divided by cos theta + sin theta is equal to secant theta into cos secant theta minus 1 divided by cosecant theta + secant theta
Answers
Cot³θ * Sin³θ /( Cosθ + Sinθ)² + Tan³θ * Cos³θ /( Cosθ + Sinθ)² = Secθ * Cosecθ - 1) /(Cosecθ + Secθ)
Step-by-step explanation:
Cot³θ * Sin³θ /( Cosθ + Sinθ)² + Tan³θ * Cos³θ /( Cosθ + Sinθ)² = Secθ * Cosecθ - 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
= ( Cosθ + Sinθ)(Cos²θ + Sin²θ - CosθSinθ)/( Cosθ + Sinθ)²
= (1 - CosθSinθ)/( Cosθ + Sinθ)
= (1 - 1/SecθCosecθ) /(1/Secθ + 1/Cosecθ)
= ( SecθCosecθ - 1)/( Cosecθ + Secθ)
= RHS
QED
proved
Cot³θ * Sin³θ /( Cosθ + Sinθ)² + Tan³θ * Cos³θ /( Cosθ + Sinθ)² = Secθ * Cosecθ - 1) /(Cosecθ + Secθ)
learn more:
Prove that: (sin3x+sinx)sinx+(cos3x-cosx)cosx=0
https://brainly.in/question/6631850
sinx-sin5x+sin9x-sin13x/cosx-cos5x -cos9x+cos13x= cot 4x
https://brainly.in/question/12448751
Answer:
here is your answer. check it...
