Question

reposting question please solve this

Answer 1

LHS = sin⁸θ-cos⁸θ
=(sin⁴θ)²-(cos⁴θ)²
=(sin⁴θ+cos⁴θ)(sin⁴θ-cos⁴θ)
={(sin²θ)²+(cos²θ)²}{(sin²θ)²-(cos²θ)²}
={(sin²θ+cos²θ)²-2sin²θcos²θ}{(sin²θ+cos²θ)(sin²θ-cos²θ)}
={(1)²-2sin²θcos²θ}{(1)(sin²θ-cos²θ)}
[∵, sin²θ+cos²θ=1]
=(sin²θ-cos²θ)(1-2sin²θcos²θ)
=RHS
∴ LHS = RHS
Hence proved.
Hope this helps you !!
# Dhruvsh

Answer 2

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