If `costheta=(p)/(sqrt(p^(2)+q^(2)))` then the value of `tantheta` is :

Options

A. a) `(q)/(sqrt(p^(2)-q^(2)))`
B. b) `(q)/(p)`
C. c) `(p)/(p^(2)+q^(2))`
D. d) `(q)/(sqrt(p^(2)+q^(2)))`

Answer

To find the value of \( \tan \theta \) given that \( \cos \theta = \frac{p}{\sqrt{p^2 + q^2}} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship of trigonometric functions**:
We know that:
\[ \cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \]
Here, we can identify the adjacent side as \( p \) and the hypotenuse as \( \sqrt{p^2 + q^2} \).

2. **Draw a right triangle**:
Let's draw a right triangle where:
- The adjacent side (base) is \( p \).
- The hypotenuse is \( \sqrt{p^2 + q^2} \).

3. **Use the Pythagorean theorem**:
According to the Pythagorean theorem:
\[ \text{Hypotenuse}^2 = \text{Adjacent}^2 + \text{Opposite}^2 \]
Plugging in the values we have:
\[ (\sqrt{p^2 + q^2})^2 = p^2 + \text{Opposite}^2 \]
This simplifies to:
\[ p^2 + q^2 = p^2 + \text{Opposite}^2 \]

4. **Solve for the opposite side**:
By subtracting \( p^2 \) from both sides, we find:
\[ q^2 = \text{Opposite}^2 \]
Taking the square root gives us:
\[ \text{Opposite} = q \]

5. **Calculate \( \tan \theta \)**:
The tangent of an angle is given by:
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{q}{p} \]

### Final Answer:
Thus, the value of \( \tan \theta \) is:
\[ \tan \theta = \frac{q}{p} \]