show that `(1+cosA+sinA)/(1-cosA+sinA) = (1+cosA)/sinA` where `A!=npi ` or `2npi-pi/2` `,ninZ` .

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`(1+sinA+cosA)/(1+sinA-cosA)`

`= ((1+sinA)+cosA)/((1+sinA)-cosA)`

`= ((1+sinA)+cosA)/((1+sinA)-cosA)xx((1+sinA)+cosA)/((1+sinA)+cosA)`

`= (1+sinA+cosA)^2/((1+sinA)^2-cos^2A)`

`= ((1+sinA)^2+2cosA(1+sinA)+cos^2A)/(2sinA(1+sinA))`

`= (1+2sinA+sin^2A+2cosA(1+sinA)+cos^2A)/(2sinA(1+sinA))`

`= (2(1+sinA)+2cosA(1+sinA))/(2sinA(1+sinA))`

`= ((1+sinA)(2+2cosA))/(2sinA(1+sinA))`

`= (2+2cosA)/(2sinA)`

`= (1+cosA)/sinA`

*So the answer is proved.*

`(1+sinA+cosA)/(1+sinA-cosA) = (1+cosA)/sinA`

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