# (sin A-cos A+1) / (sin A +cos A-1)=cos A/(1-sin A) and also........1+sin A/ cos A=cos A/1-sin A

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To do this we need to know the following;

`sin^A+cos^2A = 1`

`(Q+P)(Q-P) = Q^2-P^2`

`(Q+P)^2 = Q^2+2PQ+P^2`

Let `cosA-1 = P` and `sinA = Q`

`(sin A-cos A+1 )/ (sin A +cos A-1)`

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

`= (Q-P)/(Q+P)`

`= (Q-P)/(Q+P)*(Q+P)/(Q+P)`

`= (Q^2-P^2)/(Q+P)^2`

`= (Q^2-P^2)/(Q^2+2PQ+P^2)`

`Q^2 = sin^2A`

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

` <br> `

` Q^2+2PQ+P^2`

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

`= 1+1+2sinAcosA-2sinA-2cosA`

`= 2(-sinA+sinAcosA-cosA+1)`

`= 2[-sinA(1-cosA)+1(1-cosA)]`

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

`= 2(1-sinA)(1-cosA)`

`Q^2-P^2`

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

`= 1-cos^A-cos^2A+2cosA-1`

`= 2cosA-2cos^2A`

`= 2cosA(1-cosA)`

`(sin A-cos A+1 )/ (sin A +cos A-1)`

`= (Q^2-P^2)/(Q^2+2PQ+P^2) `

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

`= cosA/(1-sinA)`

**Therefore ;**

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

**Sources:**