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Question
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
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Solution
L.H.S. = `(cotA + cosecA - 1)/(cotA - cosecA + 1)`
= `(cotA + cosecA - (cosec^2A - cot^2A))/(cotA - cosecA + 1)` ...[cosec2A – cot2A = 1]
= `(cotA + cosecA - [(cosecA - cotA)(cosecA + cotA)])/(cotA - cosecA + 1`
= `(cotA + cosecA[1 - cosecA + cotA])/(cotA - cosecA + 1)`
= cot A + cosec A
= `cosA/sinA + 1/sinA`
= `(1 + cosA)/sinA`
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If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
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