Prove that
sin 8-cos8+1
sin 8+cos - sace-tan
using the identity sees = 1+tan's
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Answer:
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)
( Hence Proved)
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