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+1 vote
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in Trigonometry by (44.9k points)
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Prove that
`(cot theta + "cosec" theta -1 )/(cot theta - "cosec" theta +1)=(1+ cos theta)/(sin theta).`

2 Answers

+1 vote
by (17.0k points)
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Best answer

Hence Proved.

+1 vote
by (43.8k points)
We have
` LHS = ((cot theta + "cosec" theta )-1 )/((cot theta - "cosec" theta +1))`
` =(("cosec" theta + cot theta)-("cosec"^(2) theta - cot^(2)theta ))/((cot theta - "cosec" theta +1)) " " [ because 1= "cosec"^(2) theta - cot^(2)theta] `
` =(("cosec" theta + cot theta)[1- ("cosec" theta - cot theta)])/((cot theta - "cosec" theta +1))`
`=(("cosec" theta + cot theta )(cot theta - "cosec" theta +1))/((cot theta - "cosec" theta +1)) `
` =("cosec" theta + cot theta)=((1)/(sin theta) +(cos theta)/(sin theta))=(1+ cos theta)/(sin theta)= RHS. `
Hence,` (cot theta + "cosec" theta - 1)/(cot theta - "cosec" theta +1)=(1+ cos theta)/(sin theta). `

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