Question

(1) 3(sin A-cosA )^4 + 6(sinA + cos A)^2 + 4(sin^6+ cos^6) = 13.​

Answer 1

Answer:

LHS=3(sinA-cosA) ^ 4 + 6 (sinA + cosA) ^ 2 + 4 (sin ^ 6A + cos ^ 6A)

= 3 (1- sin2A) ^ 2 + 6 (1 + sin2A) +4 [(sin ^ 2x + cos ^ 2x) ^ 3 -3sin ^ 2xcos ^ 2x (sin ^ 2x + cos ^ 2x)) = 3 (1 + sin ^ 2 (2A) - 2sin2A) + 6 + 6sin2A +4 [1 - 3sin ^ 2 (2A) / 4]

= 13 + 3sin ^ 2 (2A) - 3sin ^ 2 (2A)

= 13=RHS

hence proved LHS=RHS