The value of `sin theta ` in terms of `tan theta ` is _______

Options

A. `(tan theta )/(sqrt(1-tan^(2) theta))`
B. `(tan^(2) theta )/(sqrt(1+tan^(2)theta))`
C. `(tan^(2) theta )/(sqrt(1-tan^(2)theta))`
D. `(tan theta )/(sqrt(1+tan^(2) theta))`

Explanation

To find the value of \( \sin \theta \) in terms of \( \tan \theta \), we can use the relationships between the trigonometric functions. Here’s a step-by-step solution:

### Step 1: Recall the definitions of sine, cosine, and tangent
We know that:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} \]
This means that:
\[ \sin \theta = \tan \theta \cdot \cos \theta \]

### Step 2: Use the Pythagorean identity
We also know from the Pythagorean identity that:
\[ \sin^2 \theta + \cos^2 \theta = 1 \]
From this, we can express \( \cos \theta \) in terms of \( \sin \theta \):
\[ \cos^2 \theta = 1 - \sin^2 \theta \]
Taking the square root gives:
\[ \cos \theta = \sqrt{1 - \sin^2 \theta} \]

### Step 3: Substitute \( \cos \theta \) in terms of \( \tan \theta \)
We know that:
\[ \cos \theta = \frac{1}{\sqrt{1 + \tan^2 \theta}} \]
This comes from the identity:
\[ \sec^2 \theta = 1 + \tan^2 \theta \quad \text{and} \quad \cos \theta = \frac{1}{\sec \theta} \]

### Step 4: Substitute \( \cos \theta \) back into the equation for \( \sin \theta \)
Now we can substitute this expression for \( \cos \theta \) back into our equation for \( \sin \theta \):
\[ \sin \theta = \tan \theta \cdot \frac{1}{\sqrt{1 + \tan^2 \theta}} \]

### Final Result
Thus, the value of \( \sin \theta \) in terms of \( \tan \theta \) is:
\[ \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} \]