If `tan theta=(q)/(p)` then find the value of `(p sintheta+qcos theta)/(p cos theta+qsintheta)`

Answer

To solve the problem, we start with the given information that \( \tan \theta = \frac{q}{p} \). We need to find the value of

\[ \frac{p \sin \theta + q \cos \theta}{p \cos \theta + q \sin \theta}. \]

### Step 1: Rewrite the expression using \( \tan \theta \)

We know that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore, we can express \( \sin \theta \) and \( \cos \theta \) in terms of \( \tan \theta \):

\[ \sin \theta = \tan \theta \cdot \cos \theta = \frac{q}{p} \cdot \cos \theta. \]

### Step 2: Substitute \( \sin \theta \) in the expression

Now, substituting \( \sin \theta \) in the original expression:

\[ \frac{p \left(\frac{q}{p} \cos \theta\right) + q \cos \theta}{p \cos \theta + q \left(\frac{q}{p} \cos \theta\right)}. \]

This simplifies to:

\[ \frac{q \cos \theta + q \cos \theta}{p \cos \theta + \frac{q^2}{p} \cos \theta}. \]

### Step 3: Factor out \( \cos \theta \)

Factoring out \( \cos \theta \) from both the numerator and the denominator gives:

\[ \frac{(q + q) \cos \theta}{\left(p + \frac{q^2}{p}\right) \cos \theta}. \]

### Step 4: Simplify the expression

Now, we can simplify this further:

\[ \frac{2q \cos \theta}{\left(p + \frac{q^2}{p}\right) \cos \theta}. \]

Since \( \cos \theta \) is common in both numerator and denominator, we can cancel it out (assuming \( \cos \theta \neq 0 \)):

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

### Step 5: Simplify the denominator

The denominator can be rewritten as:

\[ p + \frac{q^2}{p} = \frac{p^2 + q^2}{p}. \]

### Step 6: Final expression

Thus, we have:

\[ \frac{2q}{\frac{p^2 + q^2}{p}} = \frac{2q \cdot p}{p^2 + q^2}. \]

So, the final answer is:

\[ \frac{2pq}{p^2 + q^2}. \]