UPSEE Solved Model Paper 6

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] -
[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.