Plzz Solve this Ques.
Attachments:
Answers
Answered by
1
Answer:
Step-by-step explanation:
Answer:
is given in attachment .Hope it helps .Any query is invited.
Attachments:
Answered by
3
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
Ask any query Miss Aanya^_^
I heartly want to be your friend
Similar questions