If `tan theta=t`, then `sin 2theta` is equal to :

Options

A. `(1)/(1+t^(2))`
B. `(2t)/(1+t^(2))`
C. `(t^(2))/(1+t)`
D. `(1+t^(2))/(1+t)`

Answer

To solve the problem, we need to find the value of \(\sin 2\theta\) given that \(\tan \theta = t\).

### Step-by-Step Solution:

1. **Understanding the relationship**:
We know that \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{t}{1}\). This means we can consider a right triangle where the opposite side is \(t\) and the adjacent side is \(1\).

2. **Finding the hypotenuse**:
Using the Pythagorean theorem, the hypotenuse \(h\) can be calculated as:
\[ h = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{t^2 + 1^2} = \sqrt{t^2 + 1} \]

3. **Calculating \(\sin \theta\)**:
The sine of angle \(\theta\) is given by:
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{t}{\sqrt{t^2 + 1}} \]

4. **Calculating \(\cos \theta\)**:
The cosine of angle \(\theta\) is given by:
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{t^2 + 1}} \]

5. **Using the double angle formula**:
The formula for \(\sin 2\theta\) is:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
Substituting the values of \(\sin \theta\) and \(\cos \theta\):
\[ \sin 2\theta = 2 \left(\frac{t}{\sqrt{t^2 + 1}}\right) \left(\frac{1}{\sqrt{t^2 + 1}}\right) \]

6. **Simplifying the expression**:
\[ \sin 2\theta = 2 \cdot \frac{t}{\sqrt{t^2 + 1}} \cdot \frac{1}{\sqrt{t^2 + 1}} = \frac{2t}{t^2 + 1} \]

### Final Answer:
Thus, \(\sin 2\theta = \frac{2t}{t^2 + 1}\).

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