Question

Prove that (1 plus cot theta plus tan theta ) ( sin theta minus cos theta ) the whole divided by sec cube theta minus cosec cube theta equals sin squared theta multiply by cos squared theta

Answer 1

Hi ...dear....
here is your answer.....n
n..b.b.b

question........Prove that (1 plus cot theta plus tan theta ) ( sin theta minus cos theta ) the whole divided by sec cube theta minus cosec cube theta equals sin squared theta multiply by cos squared theta.........
....this question can be solved by changing all the terms in form of sinthetha and costhetha....
.
..see picture...
Hope it helped you!!
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#shubhendu

Answer 2

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