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Question
Prove the following identities.
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
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Solution
`(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
L.H.S = `(cot theta - cos theta)/(cot theta + cos theta)`
= `cos theta/sin theta - cos theta ÷ cos theta/sin theta + cos theta`
= `(cos theta - sin theta cos theta)/sin theta ÷ (cos theta + sin theta cos theta)/sin theta`
= `(cos theta(1 - sin theta))/sin theta ÷ (cos theta(1 + sin theta))/sin theta`
= `(cos theta(1 - sin theta))/sin theta xx sin theta/(cos theta(1 + sin theta))`
= `(1 - sin theta)/(1 + sin theta)`
R.H.S = `("cosec" - 1)/("cosec"+1)`
= `1/sin theta - 1 ÷ 1/sin theta+ 1`
= `(1 - sin theta)/sin theta ÷ (1 + sin theta)/sin theta`
= `(1 - sin theta)/sin theta xx sin theta/(1 + sin theta)`
= `(1 - sin theta)/(1 + sin theta)`
R.H.S = L.H.S ⇒ `(cot theta - cos theta)/(cot theta + cos theta) = ("cosec" theta - 1)/("cosec" theta + 1)`
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