If `x=sin^(-1)((2t)/(1+t^2))` and `y=tan^(-1)((2t)/(1-t^2))` , `-1

Explanation

To prove that \(\frac{dy}{dx} = 1\) given \(x = \sin^{-1}\left(\frac{2t}{1+t^2}\right)\) and \(y = \tan^{-1}\left(\frac{2t}{1-t^2}\right)\), we will use the substitution \(t = \tan(\theta)\).

### Step-by-step Solution:

1. **Substitution**: Let \(t = \tan(\theta)\). Then, we can express \(x\) and \(y\) in terms of \(\theta\):
\[ x = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \]
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