Math, asked by tsetandolma1212, 9 months ago

Tan^3a/1+tan^2a+cot^3a/1+cot^2a

Answers

Answered by sandy1816
7

Answer:

your answer attached in the photo

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Answered by guptavishrut
1

Answer:

Step-by-step explanation:

LHS = Tan^3A / ( 1+ Tan^2A) + Cot^3A / (1 + Cot^2a)  

= Tan^3A / Sec^2A + Cot^3A / Cosec^2A  

= (sin^3A/cos^3A) / (1 / Cos^2A) + (Cos^3A/Sin^3A) / (1 / Sin^2A)  

= Sin^3A/CosA + Cos^3A/SinA  

= (Sin^4A + Cos^4A) / SinA.CosA  

= [ (Sin^2A + Cos^2A)^2 - 2Sin^2A.Cos^2A] / SinA.CosA  

= ( 1- 2Sin^A.Cos^A)/ SinA.CosA  

RHS = SecA CosecA - 2sinAcosA  

= 1/CosA . 1/SinA - 2SinACosA  

= (1 - Sin^2A.Cos^2A) / sinAcosA  

Hence LHS = RHS (PROVED)

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