Given : q tan A = p, find the value of :
`(p sin A -q cos A)/(p sin A+q cos A)`

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{p \sin A - q \cos A}{p \sin A + q \cos A} \]

Given that \( q \tan A = p \), we can express \(\tan A\) in terms of \(p\) and \(q\):

\[ \tan A = \frac{p}{q} \]

### Step 1: Find \(\sin A\) and \(\cos A\)

Using the identity \(\tan A = \frac{\sin A}{\cos A}\), we can write:

\[ \frac{\sin A}{\cos A} = \frac{p}{q} \]

From this, we can express \(\sin A\) in terms of \(\cos A\):

\[ \sin A = \frac{p}{q} \cos A \]

### Step 2: Use the Pythagorean Identity

We know from the Pythagorean identity that:

\[ \sin^2 A + \cos^2 A = 1 \]

Substituting \(\sin A\) from the previous step:

\[ \left(\frac{p}{q} \cos A\right)^2 + \cos^2 A = 1 \]

This simplifies to:

\[ \frac{p^2}{q^2} \cos^2 A + \cos^2 A = 1 \]

Factoring out \(\cos^2 A\):

\[ \cos^2 A \left(\frac{p^2}{q^2} + 1\right) = 1 \]

### Step 3: Solve for \(\cos^2 A\)

This gives us:

\[ \cos^2 A = \frac{1}{\frac{p^2}{q^2} + 1} = \frac{1}{\frac{p^2 + q^2}{q^2}} = \frac{q^2}{p^2 + q^2} \]

Taking the square root, we find:

\[ \cos A = \frac{q}{\sqrt{p^2 + q^2}} \]

### Step 4: Find \(\sin A\)

Now substituting back to find \(\sin A\):

\[ \sin A = \frac{p}{q} \cos A = \frac{p}{q} \cdot \frac{q}{\sqrt{p^2 + q^2}} = \frac{p}{\sqrt{p^2 + q^2}} \]

### Step 5: Substitute \(\sin A\) and \(\cos A\) into the Expression

Now we substitute \(\sin A\) and \(\cos A\) into the original expression:

\[ \frac{p \cdot \frac{p}{\sqrt{p^2 + q^2}} - q \cdot \frac{q}{\sqrt{p^2 + q^2}}}{p \cdot \frac{p}{\sqrt{p^2 + q^2}} + q \cdot \frac{q}{\sqrt{p^2 + q^2}}} \]

This simplifies to:

\[ \frac{\frac{p^2 - q^2}{\sqrt{p^2 + q^2}}}{\frac{p^2 + q^2}{\sqrt{p^2 + q^2}}} \]

### Step 6: Simplify the Expression

The \(\sqrt{p^2 + q^2}\) cancels out:

\[ \frac{p^2 - q^2}{p^2 + q^2} \]

Thus, the final answer is:

\[ \frac{p^2 - q^2}{p^2 + q^2} \]