Trigonometric Equation Analysis

We are given the equations:

• \(p + q \cot \theta = 3 \cosec \theta\) (1)
• \(q - p \cot \theta = 2 \cosec \theta\) (2)

The objective is to find \(\tan \theta\).

Eliminating \(\cosec \theta\)

Multiply Equation (1) by 2 and Equation (2) by 3 to standardize the \(\cosec \theta\) term on the right side:

• \(2 \times (p + q \cot \theta) = 2 \times 3 \cosec \theta \implies 2p + 2q \cot \theta = 6 \cosec \theta\)
• \(3 \times (q - p \cot \theta) = 3 \times 2 \cosec \theta \implies 3q - 3p \cot \theta = 6 \cosec \theta\)

Equating the left-hand sides since the right-hand sides are equal:

\(2p + 2q \cot \theta = 3q - 3p \cot \theta\)

Solving for \(\cot \theta\)

Rearrange the equation to group \(\cot \theta\) terms on one side and constants on the other:

\(2q \cot \theta + 3p \cot \theta = 3q - 2p\)

Factor out \(\cot \theta\):

\(\cot \theta (2q + 3p) = 3q - 2p\)

Isolate \(\cot \theta\):

\(\cot \theta = \frac{3q - 2p}{2q + 3p}\)

Calculating \(\tan \theta\)

Using the reciprocal identity \(\tan \theta = \frac{1}{\cot \theta}\):

\(\tan \theta = \frac{1}{\frac{3q - 2p}{2q + 3p}}\)

\(\tan \theta = \frac{2q + 3p}{3q - 2p}\)

The final answer is \(\boxed{\frac{(2q+3p)}{(3q+2p)}}\)