If ` x = cos^(-1) "" (1)/(sqrt"" (t^2 +1)). Y= sin^(-1)"" (t )/(sqrt""(1+t^2))` then ` (dy)/(dx)` in independent of t .

Answer

To solve the problem, we need to find \(\frac{dy}{dx}\) given the equations:

\[ x = \cos^{-1}\left(\frac{1}{\sqrt{t^2 + 1}}\right) \]
\[ y = \sin^{-1}\left(\frac{t}{\sqrt{1 + t^2}}\right) \]

We will use the parametric differentiation method to find \(\frac{dy}{dx}\).

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

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

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

Using the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\), we have:

\[ \sqrt{t^2 + 1} = \sqrt{\tan^2(\theta) + 1} = \sqrt{\sec^2(\theta)} = \sec(\theta) \]

Thus, we can rewrite \(x\):

\[ x = \cos^{-1}\left(\frac{1}{\sec(\theta)}\right) = \cos^{-1}(\cos(\theta)) = \theta \]

### Step 3: Simplify \(y\)

Now, for \(y\):

\[ \sqrt{1 + t^2} = \sqrt{1 + \tan^2(\theta)} = \sqrt{\sec^2(\theta)} = \sec(\theta) \]

So we can rewrite \(y\):

\[ y = \sin^{-1}\left(\frac{\tan(\theta)}{\sec(\theta)}\right) = \sin^{-1}(\sin(\theta)) = \theta \]

### Step 4: Express \(x\) and \(y\) in terms of \(\theta\)

Now we have:

\[ x = \theta \]
\[ y = \theta \]

### Step 5: Find \(\frac{dy}{dx}\)

Now, we can find \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

Since both \(y\) and \(x\) are equal to \(\theta\), we differentiate:

\[ \frac{dy}{d\theta} = 1 \]
\[ \frac{dx}{d\theta} = 1 \]

Thus, we have:

\[ \frac{dy}{dx} = \frac{1}{1} = 1 \]

### Conclusion

Since \(\frac{dy}{dx} = 1\), it is independent of \(t\).