Question

( a sin A + b cos A)^2 + (a cos A – b sin A)^2 = a^2 + b^2

Answer 1

Answer:

this is the answer to the question

Answer 2

(a sin A +b cos A) ^2+ (a cos A- b sin A)^2= a^2 + b^2

by identity (a+b)^2

a^2 sin^2A +b^2 cos^2A + 2ab sinA cosA + a^2 cos^2A + b^2 sin^2A - 2ab sinA cosA = a^2 + b^2

[a^2 sin^2A + a^2cos^2A]+ [b^2 cos^2A+ b^2 sin^2A]+ [ 2ab sinA cosA -2ab sinA cos A] = a^2+ b^2

a^2(sin^2A+ cos ^2A) + b^2(sin^2A + cos^2A) = a^2 +b^2

by identity sin^2A+ cos^A= 1

a^2 + b^2 = a^2 + b^2

LHS =RHS

hence proved

Hope you will understand....