Question

Plzzzzzz answer this question I WILL MARK AS BRAINLIEST

Answer 1

Step-by-step explanation:

I hope you know that sin²θ+cos²θ=1

LHS=2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1

=2[(sin²θ)³+(cos²θ)³] - 3[(sin²θ)²+(cos²θ)²] + 1

=2(sin²θ+cos²θ)(sin⁴θ+cos⁴θ-sin²θcos²θ) - 3[(sin²θ+cos²θ)²-2sin²θcos²θ] + 1

=2[(sin²θ+cos²θ)²-3sin²θcos²θ] - 3(1-2sin²θcos²θ) + 1

=2(1-3sin²θcos²θ) - 3(1-2sin²θcos²θ) + 1

=2 - 6sin²θcos²θ - 3 + 6sin²θcos²θ + 1

=0=RHS

HENCE PROVED