Question

(cos ⁶theta +sin⁶theta )+3sin²theta cos²theta​

Answer 1

Step-by-step explanation:

Answer

L.H.S. =sin⁶θ+cos⁶θ=(sin²θ)³+(cos²θ)³

Assume sin²θ=a and cos²θ=b

∴ L.H.S. =a³ + b³

=(a+b)³ −2ab(a+b) =(sin²θ−cos²θ−3sin²θcos²θ(sin²θ+cos²θ)

=(1)³ −3sin²θcos²θ[∵sin²θ+cos²θ=1]

=1−3sin²θcos²θ

= R.H.S.

hence proved

Answer 2

$({cos}^{6} \theta + {sin}^{6} \theta) + 3 {sin}^{2} \theta \: {cos}^{2} \theta \\ = ( {cos}^{2} \theta) ^{3} + ( { {sin}^{2} \theta})^{3} + 3 {sin}^{2} \theta \: {cos}^{2} \theta \\ = ( {cos}^{2} \theta + {sin}^{2} \theta)( {cos}^{4} \theta + {sin}^{4} \theta - {sin}^{2} \theta {cos}^{2} \theta) + 3 {sin}^{2} \theta {cos}^{2} \theta \\ = {cos}^{4} \theta + {sin}^{4} \theta + 2 {sin}^{2} \theta {cos}^{2} \theta \\ = ( { {sin}^{2} \theta + {cos}^{2} \theta})^{2} \\ = 1$