If `tantheta =p/q`, then the value of `(p sintheta-q costheta)/(p sintheta+q costheta)` is

Options

A. `(p^(2)-q^(2))/(p^(2)+q^(2))`
B. `(p^(2)+q^(2))/(p^(2)-q^(2))`
C. 0
D. 1

Answer

To solve the problem, we need to find the value of the expression \(\frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta}\) given that \(\tan \theta = \frac{p}{q}\).

### Step-by-Step Solution:

1. **Start with the given information**:
\[ \tan \theta = \frac{p}{q} \]

2. **Express \(\sin \theta\) and \(\cos \theta\)** in terms of \(p\) and \(q\):
From the definition of tangent, we have:
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{p}{q} \]
This implies:
\[ \sin \theta = \frac{p}{\sqrt{p^2 + q^2}} \quad \text{and} \quad \cos \theta = \frac{q}{\sqrt{p^2 + q^2}} \]

3. **Substitute \(\sin \theta\) and \(\cos \theta\)** into the expression:
Substitute the values of \(\sin \theta\) and \(\cos \theta\) into the expression:
\[ \frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta} = \frac{p \left(\frac{p}{\sqrt{p^2 + q^2}}\right) - q \left(\frac{q}{\sqrt{p^2 + q^2}}\right)}{p \left(\frac{p}{\sqrt{p^2 + q^2}}\right) + q \left(\frac{q}{\sqrt{p^2 + q^2}}\right)} \]

4. **Simplify the numerator and denominator**:
The numerator becomes:
\[ \frac{p^2 - q^2}{\sqrt{p^2 + q^2}} \]
The denominator becomes:
\[ \frac{p^2 + q^2}{\sqrt{p^2 + q^2}} \]

5. **Combine the fractions**:
Now, we can write the expression as:
\[ \frac{p^2 - q^2}{p^2 + q^2} \]

6. **Final Result**:
Thus, the value of the expression \(\frac{p \sin \theta - q \cos \theta}{p \sin \theta + q \cos \theta}\) is:
\[ \frac{p^2 - q^2}{p^2 + q^2} \]