`sinx = (2t)/(1+t^(2)), tan y = (2t)/(1-t^(2))`

Explanation

To solve the problem given \( \sin x = \frac{2t}{1+t^2} \) and \( \tan y = \frac{2t}{1-t^2} \), we need to find \( \frac{dy}{dx} \). We will differentiate both equations with respect to \( t \) and then use the chain rule to find \( \frac{dy}{dx} \).

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

Given:
\[ \sin x = \frac{2t}{1+t^2} \]

Using the chain rule:
\[ \frac{d}{dt}(\sin x) = \cos x \cdot \frac{dx}{dt} \]

Now, differentiate the right-hand side using the quotient rule:
\[ \frac{d}{dt}\left(\frac{2t}{1+t^2}\right) = \frac{(1+t^2)(2) - (2t)(2t)}{(1+t^2)^2} \]
\[ = \frac{2(1+t^2) - 4t^2}{(1+t^2)^2} = \frac{2 - 2t^2}{(1+t^2)^2} \]

Setting the derivatives equal:
\[ \cos x \cdot \frac{dx}{dt} = \frac{2(1-t^2)}{(1+t^2)^2} \]

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

Given:
\[ \tan y = \frac{2t}{1-t^2} \]

Using the chain rule:
\[ \frac{d}{dt}(\tan y) = \sec^2 y \cdot \frac{dy}{dt} \]

Now, differentiate the right-hand side using the quotient rule:
\[ \frac{d}{dt}\left(\frac{2t}{1-t^2}\right) = \frac{(1-t^2)(2) - (2t)(-2t)}{(1-t^2)^2} \]
\[ = \frac{2(1-t^2) + 4t^2}{(1-t^2)^2} = \frac{2 + 2t^2}{(1-t^2)^2} \]

Setting the derivatives equal:
\[ \sec^2 y \cdot \frac{dy}{dt} = \frac{2(1+t^2)}{(1-t^2)^2} \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

From the previous steps, we have:
\[ \frac{dy}{dt} = \frac{2(1+t^2)}{(1-t^2)^2 \sec^2 y} \]
\[ \frac{dx}{dt} = \frac{2(1-t^2)}{(1+t^2)^2 \cos x} \]

Thus:
\[ \frac{dy}{dx} = \frac{\frac{2(1+t^2)}{(1-t^2)^2 \sec^2 y}}{\frac{2(1-t^2)}{(1+t^2)^2 \cos x}} \]

This simplifies to:
\[ \frac{dy}{dx} = \frac{(1+t^2) \cdot (1+t^2)^2 \cos x}{(1-t^2)^2 \cdot (1-t^2)} \]

### Step 4: Final Simplification

After simplifying, we can write:
\[ \frac{dy}{dx} = \frac{(1+t^2)^3 \cos x}{(1-t^2)^3} \]