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Question
Prove the following identities:
`((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA)) = 2cotA`
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Solution
L.H.S. = `((cosecA - cotA)^2 + 1)/(secA(cosecA - cotA))`
= `(cosec^2A + cot^2A - 2cosecAcotA + 1)/(secA(cosecA - cotA))`
= `(cosec^2A + (1 + cot^2A) - 2cosecAcotA)/(secA(cosecA - cotA))`
= `(cosec^2A + cosec^2A - 2cosecAcotA)/(secA(cosecA - cotA))`
= `(2cosec^2A - 2cosecAcotA)/(secA(cosecA - cotA))`
= `(2cosecA(cosecA - cotA))/(secA(cosecA - cotA))`
= `(2cosecA)/secA`
= `(2 1/sinA)/(1/cosA)`
= `2/sinA xx cosA/1`
= `2 cosA/sinA`
= 2 cot A = R.H.S.
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= `cosθ/sinθ + square/cosθ`
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∴ L.H.S. = R.H.S.
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