If `y = cos^(-1) ((1)/( sqrt(1+t^(2)))), x = sin^(-1) (sqrt((t^(2))/(1 + t^(2)))), "find " (dy)/(dx)`

Answer

To find \(\frac{dy}{dx}\) given the equations \(y = \cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)\) and \(x = \sin^{-1}\left(\sqrt{\frac{t^2}{1+t^2}}\right)\), we will use the chain rule. We will differentiate \(y\) and \(x\) with respect to \(t\) and then use the relationship:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

### Step 1: Differentiate \(y\) with respect to \(t\)

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

To differentiate \(y\), we will use the substitution \(t = \tan(\theta)\). Therefore, we have:

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

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

From the substitution, we know:
\[ \theta = \tan^{-1}(t) \]

Now, differentiating \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = \frac{d}{dt}(\tan^{-1}(t)) = \frac{1}{1+t^2} \]

### Step 2: Differentiate \(x\) with respect to \(t\)

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

Using the same substitution \(t = \tan(\theta)\):
\[ \sqrt{\frac{t^2}{1+t^2}} = \sqrt{\frac{\tan^2(\theta)}{\sec^2(\theta)}} = \sqrt{\sin^2(\theta)} = \sin(\theta) \]

Thus:
\[ x = \sin^{-1}(\sin(\theta)) = \theta \]

Again, we have:
\[ \theta = \tan^{-1}(t) \]

Now, differentiating \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = \frac{d}{dt}(\tan^{-1}(t)) = \frac{1}{1+t^2} \]

### Step 3: Calculate \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{1}{1+t^2}}{\frac{1}{1+t^2}} = 1 \]

### Final Answer:
\[ \frac{dy}{dx} = 1 \]

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