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.
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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
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