We can solve this problem using basic formula of trigonometric functions Basic formula sin2A+cos2A = 1 By using this formula we can find the sum of the given trigonometric function We have 9sin4θ−6sin6θ+9cos4θ−6cos6θ = 9(sin4θ+cos4θ) - 6(sin6θ+cos6θ) = 9[(cos2θ+sin2θ)2−2sin2θcos2θ] - 6[(sin2θ)3+(cos2θ)3] [Since, a4+b4 = (a2+b2)2−2a2b2] = 9[1−2sin2θcos2θ] -
6[(cos2θ+sin2θ)((cos2θ+sin2θ)2−3cos2θsin2θ)]
[since, sin2θ+cos2θ = 1 ; a3+b3 = (a + b) ((a+b)2−3ab)] = 9[1−2sin2θcos2θ] - 6[1−3sin2θcos2θ] = 9 - 18sin2θcos2θ - 6 + 18sin2θcos2θ = 9 - 6 = 3 Therefore , The sum of 9sin4θ−6sin6θ and 9cos4θ−6cos6θ is 3.