Question

prove that 2(Sin⁶θ+Cos⁶θ)-3(Sin⁴θ+Cos⁴θ)+1 = 0.


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Answer 1

Answer:

LHS = RHS

Hence , Proved

Step-by-step explanation:

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Answer 2

$\huge\underline\mathfrak\purple{Answer}$

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)

= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}

=2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}

=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)

=2-6sin²θcos²θ-3+6sin²θcos²θ

=-1

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