Question

1 + cot theta + tan theta into sin theta minus cos theta upon sec cube theta minus cos cube theta = 2 sin square theta into cos2 theta.

Answer 1

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Answer 2

Step-by-step explanation:

Here we want to prove

(1+cotθ+tanθ)(sinθ-cosθ) / sec³θ-cosec³θ   =  sin²θcos²θ

L.H.S⇒ {1+cosθ/sinθ+sinθ/cosθ}  (sinθ-cosθ)  /    (1/cos³θ) - (1/sin³θ)

⇒{(sinθcosθ+cos²θ+sin²θ) / sinθcosθ} (sinθ-cosθ) /

    (sin³θ-cos³θ)/cos³θsin³θ

⇒{(sinθcosθ+1) / sinθcosθ} (sinθ-cosθ) /

   (sinθ- cosθ)(sin²θ+sinθcosθ+cos²θ)/ sinθcosθ.sin²θcos²θ

⇒(sinθcosθ+1)(sinθ-cosθ) /   [(sinθ-cosθ)(sinθcosθ+1)/sin²θcos²θ}

⇒sin²θcos²θ     ∵a/b÷c/d= ad/bc

⇒R.H.S