If ` tan x = (2t)/(1 -t^(2)) " and sin y" = (2t)/(1 + t^(2))` , then the value of
` (dy)/(dx)` is

Options

A. 1
B. t
C. `(1)/(1-t)`
D. `(1)/(1 + t)`

Explanation

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

### Step 1: Express \(x\) and \(y\) in terms of \(t\)

From the given equations, we can express \(x\) and \(y\) in terms of \(t\).

1. **For \(x\)**:
\[ \tan x = \frac{2t}{1 - t^2} \]
Using the identity for the tangent of a double angle, we have:
\[ \tan x = \tan(2\theta) \quad \text{where } t = \tan(\theta) \]
Thus, we can write:
\[ x = 2\theta = 2\tan^{-1}(t) \]

2. **For \(y\)**:
\[ \sin y = \frac{2t}{1 + t^2} \]
Using the identity for the sine of a double angle, we have:
\[ \sin y = \sin(2\theta) \quad \text{where } t = \tan(\theta) \]
Thus, we can write:
\[ y = 2\theta = 2\tan^{-1}(t) \]

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

Now we differentiate both \(x\) and \(y\) with respect to \(t\).

1. **Differentiate \(x\)**:
\[ \frac{dx}{dt} = \frac{d}{dt}(2\tan^{-1}(t)) = 2 \cdot \frac{1}{1 + t^2} \]

2. **Differentiate \(y\)**:
\[ \frac{dy}{dt} = \frac{d}{dt}(2\tan^{-1}(t)) = 2 \cdot \frac{1}{1 + t^2} \]

### Step 3: Use the chain rule to find \(\frac{dy}{dx}\)

Using the chain rule, we can express \(\frac{dy}{dx}\) as follows:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the values we found:
\[ \frac{dy}{dx} = \frac{2/(1 + t^2)}{2/(1 + t^2)} = 1 \]

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

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