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

Answer

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

### Step 1: Express \( x \) and \( y \) in terms of \( t \)
From the given equations, we can relate \( x \) and \( y \) to \( t \).

1. **For \( \sin x = \frac{2t}{1+t^2} \)**:
This can be recognized as the sine double angle formula:
\[ \sin x = \sin(2\theta) \quad \text{where } t = \tan(\theta) \]
Thus, we can write:
\[ x = 2\theta \]

2. **For \( \tan y = \frac{2t}{1-t^2} \)**:
This can also be recognized as the tangent double angle formula:
\[ \tan y = \tan(2\theta) \]
Therefore:
\[ y = 2\theta \]

### Step 2: Relate \( x \) and \( y \)
From the above expressions, we see that:
\[ x = 2\theta \quad \text{and} \quad y = 2\theta \]
This implies:
\[ x = y \]

### Step 3: Differentiate both sides
Now, we differentiate both sides with respect to \( t \):
\[ \frac{dx}{dt} = \frac{dy}{dt} \]

### Step 4: Find \( \frac{dy}{dx} \)
Using the fact that \( x = y \), we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Since \( \frac{dx}{dt} = \frac{dy}{dt} \), we can conclude:
\[ \frac{dy}{dx} = 1 \]

### Final Result
Thus, the value of \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = 1 \]