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Question
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA ` using the identity cosec2 A = 1 cot2 A.
Solution
`(cos A-sinA+1)/(cosA+sinA-1)=cosecA+cotA`
Using the identity cosec2A = 1 + cot2A,
L.H.S = `(cos A-sinA+1)/(cosA+sinA-1)`
= `(cosA/sinA-sinA/sinA+1/sinA)/(cosA/sinA+sinA/sinA+1/sinA)`
= `(cotA-1+cosec A)/(cotA+1-cosec A)`
= `({(cotA)-(1-cosec A)}{(cotA)-(1-cosec A)})/({(cotA)+(1-cosec A)}{(cotA)-(1-cosec A)})`
= `(cot A - 1 + cosecA)^2/((cotA)^2-(1-cosecA)^2)`
= `(cot^2A+1+cosec^2A-2cotA-2cosec A+2cotAcosec A)/(cot^2A-(1+cosec^2 A-2cosec A))`
= `(2cosec^2 A+2cotAcosec A-2cotA-2cosec A)/(cot^2A-1-1cosec^2 A+2cosec A)`
= `(2cosec A(cosecA+cotA)-2(cotA+cosec A))/(cot^2A-cosec^2A-1+2cosec A)`
= `((cosec A+cotA)(2cosec A-2))/(-1-1+2cosec A)`
= `((cosec A+cotA)(2cosec A-2))/(2cosec A-2)`
= cosec A + cot A
= R.H.S
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