10. Prove that 2(sin ^6teta + cos^6teta)-3(sin*4 teta+ cos* 4teta)+1=0.
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Step-by-step explanation:
2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1
= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1
= 2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1
= 2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1
= 2-6sin²θcos²θ-3+6sin²θcos²θ+1
= -1+1
=0
Hope it helps!Brainliest it sister!
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