Question

cot3A-sin3A/(cosA+sinA)2+tan3A×cos3A/(cos+sin)2=sec×cosec-1/cosecA+secA​

Answer 1

Answer:

Cot³ASin³A /(CosA + SinA)²  + Tan³A.Cos³A/(CosA + SinA)² = (SecA.CosecA - 1)/(CosecA + SecA)

LHS = Cot³ASin³A /(CosA + SinA)²  + Tan³A.Cos³A/(CosA + SinA)²

= Cos³A /(CosA + SinA)²  + Sin³A/(CosA + SinA)²

= (Cos³A  + Sin³A)/(CosA + SinA)²

using a³ + b³ = (a + b)(a² + b² - ab)

= (CosA + SinA)(Cos²A  + Sin²A - CosASinA)/(CosA + SinA)²  

= (Cos²A  + Sin²A - CosASinA)/(CosA + SinA)

= (1 - CosASinA)/(CosA + SinA)

DIviding numerator & Denominator by CosASinA

= (1/CosASinA  - 1)/(1/SinA  + 1/CosA)

= (SecACosecA - 1)/(CosecA + SecA)

= RHS

QED

Proved

Cot³ASin³A /(CosA + SinA)²  + Tan³A.Cos³A/(CosA + SinA)² = (SecA.CosecA - 1)/(CosecA + SecA)

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