# How do i solve this trigonometic proof while only manipulating ONE side? (sinx-cosx) / (sinx+cosx) = (2sin^2x-1) / (1+2sinxcosx)

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`(sinx-cosx)/(sinx+cosx) = (2sin^2 x -1)/(1+2sinxcosx)`

`` We will start from the left side to get to the right side.

`==gt (sinx-cosx)/(sinx+cosx)```

We will multiply and divide by (sinx+cosx)

`==gt ((sinx-cosx)(sinx+cosx))/((sinx+cosx)^2 `

`==gt (sin^2 x - cos^2 x)/(sin^2 x + 2sinxcosx + cos^2 x)`

`` Now we will use the following identity:

We know that: `sin^2 x + cos^2 x = 1`

`` `==gt cos^2 x = 1-sin^2 x `

`==gt (sin^2 x - (1-sin^2 x))/(1+2sinxcosx) `

`==gt (sin^2 x - 1 +sin^2 x)/(1+2sinxcosx) `

`==gt (2sin^2 x -1)/(1+2sinxcosx)` " which is the right side"