Question

tan^3A/1+tan^2A + cot^3A/1-cot^2A = secA×cosecA - 2sinA×cosA​

Answer 1

Step-by-step explanation:

SOLUTION:-

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)

Answer 2

Answer:  (tan^3A/1+tan^2A) +(Cot^3A/1+Cot^2A) = SecA.CosecA - 2Sina.CosA

Step-by-step explanation: mark me as brainliest

Thank you