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If sinx=(2t)/(1+t^2),tany=(2t)/(1-t^2),"...

If `sinx=(2t)/(1+t^2),tany=(2t)/(1-t^2),"f i n d"(dy)/(dx)dot`

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Let `t = tan theta`
Then, `sinx = (2tan theta)/(1+tan^2theta)`
As, `(2tan theta)/(1+tan^2theta) = sin2theta`,
`=> sin x = sin2theta`
`=> x= 2npi+2theta`
`=>dx/ (d theta) = 2 `
Now, `tan y = (2tan theta)/(1-tan^2theta)`
As, `(2tan theta)/(1- tan^2theta) = tan2theta`,
...
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