Question

what solve this Q..........​

Answer 1

Answer:

sin^6θ+3sin^5θ+3sin^4θ+sin^3θ+2sin^2θ+2sinθ-2

Step-by-step explanation:

sinθ=1-sin^2θ

sinθ=cos^2θ

Since,

sin^2θ+cos^2θ=1

cos^12θ+3cos^10θ+3cos^8θ+cos^6θ+2cos^4θ+2cos^2θ-2

cos^12θ= sin^6θ

cos^10θ= sin^5θ

cos^8θ= sin^4θ

cos^6θ= sin^3θ

cos^4θ= sin^2θ

cos^2θ= sin^θ

sin^6θ+3sin^5θ+3sin^4θ+sin^3θ+2sin^2θ+2sinθ-2

Answer 2

Answer:

heya....

sin theta + sin^2 theta = 1

sin A = 1 - sin^2A

sin A = cos^2 A

cos^12 + 3cos^10 + 3cos^8 + 2cos^4 + 2 cos^2 - 2

= (cos^2 A)^6 + 3(coa^2 A )^5 + 3(cos^2A)^4 + 2( cos^2A)^2 + 2 cos^2 A - 2

[ sin = cos^2]

sjn^6 A + 3sin ^5A + 3sin^4 A + 2sin^2A + 2 sin A - 2

hope it help u my ch friend ✌✌✌✌