If `x=sin^-1""(2t)/(1+t^2)` and
`y=tan ^-1""(2t)/(1-t) " the n find " dy/dx.`

Explanation

To find \( \frac{dy}{dx} \) given the equations \( x = \sin^{-1} \left( \frac{2t}{1+t^2} \right) \) and \( y = \tan^{-1} \left( \frac{2t}{1-t} \right) \), we can follow these steps:

### Step 1: Substitute \( t = \tan \theta \)

Let \( t = \tan \theta \). Then, we can rewrite \( x \) and \( y \) in terms of \( \theta \).

### Step 2: Rewrite \( x \)

...