Question

Prove that cos cube theta + sin cube theta/cos theta + sin theta added to cos cube theta - sin cube theta/cos theta - sin theta = 2.

Answer 1

(cos³θ+sin³θ/cosθ+sinθ)+(cos³θ-sin³θ/cosθ-sinθ)
={(cosθ+sinθ)³-3cosθsinθ(cosθ+sinθ)}/(cosθ+sinθ)+{(cosθ-sinθ)³+3cosθsinθ(cosθ-sinθ)}/(cosθ-sinθ)
=(cosθ+sinθ){(cosθ+sinθ)²-3cosθsinθ}/(cosθ+sinθ)+(cosθ-sinθ){(cosθ-sinθ)²+3cosθsinθ}/(cosθ-sinθ)
=(cosθ+sinθ)²-3cosθsinθ+(cosθ-sinθ)²+3cosθsinθ
=cos²θ+2cosθsinθ+sin²θ+cos²θ-2cosθsinθ+sin²θ
=1+1
=2 (Proved)

Answer 2

Answer:

Step -by-step explanation:

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