If `sinx=(2t)/(1+t^2)` , `tany=(2t)/(1-t^2)` , find `(dy)/(dx)` .

Explanation

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

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

We start with the equation:
\[ \sin x = \frac{2t}{1+t^2} \]
Differentiating both sides with respect to \(t\):
\[ \frac{d}{dt}(\sin x) = \frac{d}{dt}\left(\frac{2t}{1+t^2}\right) \]
Using the chain rule on the left side:
\[ \cos x \frac{dx}{dt} = \frac{(1+t^2)(2) - 2t(2t)}{(1+t^2)^2} \]
This simplifies to:
\[ \cos x \frac{dx}{dt} = \frac{2 + 2t^2 - 4t^2}{(1+t^2)^2} = \frac{2 - 2t^2}{(1+t^2)^2} \]

### Step 2: Solve for \(\frac{dx}{dt}\)

Rearranging the equation gives:
\[ \frac{dx}{dt} = \frac{2 - 2t^2}{(1+t^2)^2 \cos x} \]

### Step 3: Find \(\cos x\)

Using the identity \(\cos^2 x + \sin^2 x = 1\):
\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{2t}{1+t^2}\right)^2 \]
Calculating \(\sin^2 x\):
\[ \sin^2 x = \frac{4t^2}{(1+t^2)^2} \]
Thus,
\[ \cos^2 x = 1 - \frac{4t^2}{(1+t^2)^2} = \frac{(1+t^2)^2 - 4t^2}{(1+t^2)^2} = \frac{1 + 2t^2 + t^4 - 4t^2}{(1+t^2)^2} = \frac{1 - 2t^2 + t^4}{(1+t^2)^2} \]
So,
\[ \cos x = \frac{\sqrt{(1 - t^2)^2}}{1+t^2} = \frac{1 - t^2}{1+t^2} \]

### Step 4: Substitute \(\cos x\) back into \(\frac{dx}{dt}\)

Substituting \(\cos x\) into the equation for \(\frac{dx}{dt}\):
\[ \frac{dx}{dt} = \frac{2 - 2t^2}{(1+t^2)^2 \cdot \frac{1 - t^2}{1+t^2}} = \frac{2(1+t^2)(1 - t^2)}{(1+t^2)^2(2 - 2t^2)} = \frac{2(1 - t^2)}{(1+t^2)(1 - t^2)} = \frac{2}{1+t^2} \]

### Step 5: Differentiate \(\tan y\) with respect to \(t\)

Now, we differentiate:
\[ \tan y = \frac{2t}{1-t^2} \]
Differentiating both sides:
\[ \sec^2 y \frac{dy}{dt} = \frac{(1-t^2)(2) - 2t(-2t)}{(1-t^2)^2} \]
This simplifies to:
\[ \sec^2 y \frac{dy}{dt} = \frac{2(1-t^2) + 4t^2}{(1-t^2)^2} = \frac{2 + 2t^2}{(1-t^2)^2} \]

### Step 6: Solve for \(\frac{dy}{dt}\)

Rearranging gives:
\[ \frac{dy}{dt} = \frac{2 + 2t^2}{(1-t^2)^2 \sec^2 y} \]

### Step 7: Find \(\sec^2 y\)

Using the identity \(\sec^2 y = 1 + \tan^2 y\):
\[ \tan^2 y = \left(\frac{2t}{1-t^2}\right)^2 = \frac{4t^2}{(1-t^2)^2} \]
Thus,
\[ \sec^2 y = 1 + \frac{4t^2}{(1-t^2)^2} = \frac{(1-t^2)^2 + 4t^2}{(1-t^2)^2} = \frac{1 - 2t^2 + t^4 + 4t^2}{(1-t^2)^2} = \frac{1 + 2t^2 + t^4}{(1-t^2)^2} \]

### Step 8: Substitute \(\sec^2 y\) back into \(\frac{dy}{dt}\)

Substituting gives:
\[ \frac{dy}{dt} = \frac{2(1+t^2)}{(1-t^2)^2} \cdot \frac{(1-t^2)^2}{1 + 2t^2 + t^4} = \frac{2(1+t^2)}{1 + 2t^2 + t^4} \]

### Step 9: Find \(\frac{dy}{dx}\)

Finally, we can find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2(1+t^2)}{1 + 2t^2 + t^4}}{\frac{2}{1+t^2}} = \frac{(1+t^2)(1+t^2)}{1 + 2t^2 + t^4} = \frac{(1+t^2)^2}{1 + 2t^2 + t^4} \]
Since \(1 + 2t^2 + t^4 = (1+t^2)^2\), we find:
\[ \frac{dy}{dx} = 1 \]

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