If `x=sin^(-1)((2t)/(1+t^2))a n dy=tan^(-1)((2t)/(1-t^2)),t > 1. P rov et h a t(dy)/(dx)=-1`

Explanation

`sin x=frac{2t}{1+t^2}`
`cosx frac{dx}{dt}=frac{(1+t^2)2-2t(0+2t)}{(1+t^2)^2}`
`=frac{2+2t^2-4t^2}{(1+t^2)^2}`
`=frac{2-2t^2}{(1+t^2)^2}`
`frac{dy}{dx}=frac{2-2t^2}{(1+t^2)^2} xx frac{1+t^2}{sqrt((1+t^2)^2-4t^2)}` Since `cosx=sqrt(1-sin^2x)`
`frac{dx}{dt}=-frac{2}{1+t^2} ---(1)`
We know that `tan y=frac{2t}{1+t^2}`
`sec^2 y frac{dy}{dt}=frac{2+2t^2}{(1+t^2)^2}`
`frac{dy}{dx}=frac{2(1+t^2)}{(1+t^2)^2} xx frac{1}{[1+[frac{2t}{1-t^2}]^2]}` Since `sec^2 x=tan^2+1`
`frac{dy}{dx}=frac{2(1-t^2)}{(1-t^2)^2}`
`frac{dy}{dt}=frac{2}{1+t^2} ---(2)`
`frac{dy}{dx}=frac{frac{dy}{dt}}{frac{dx}{dt}}=-1`