Question

. Prove that 2 (sin6A + cos6A) – 3 (sin4A + cos4A ) + 1 = 0

Answer 1

Step-by-step explanation:

LHS=2(sin6θ+cos6θ)−3(sin4θ+cos4θ)+1

=2{(sin2θ+cos2θ)3−3sin2θcos2θ(sin2θ+cos2θ)}−3(sin2θ+cos2θ)2−2(sin2θcos2θ)}+1

We know, [sin²x+cos²x=1]

=2{1−3sin2θcos2θ}−3{1−2sin2θcos2θ}+1

=2−6sin2θcos2θ−3+6sin2θcos2θ+1=0

=RHS