Expression Setup

We need to find the value of the expression \(\left(p+\frac{1}{q}\right)\left(q+\frac{1}{p}\right)\). The given expressions for \(p\) and \(q\) are:

\(p = \frac{\sin \theta}{1 + \cos \theta + \sin \theta}\)

\(q = \frac{1 + \sin \theta}{1 + \sin \theta - \cos \theta}\)

Simplify p

We use the substitution \(t = \tan\left(\frac{\theta}{2}\right)\). This implies: \(\sin \theta = \frac{2t}{1+t^2}\) and \(\cos \theta = \frac{1-t^2}{1+t^2}\).

Substituting these into the expression for \(p\): \(p = \frac{\frac{2t}{1+t^2}}{1 + \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2}}\)

To simplify, multiply the numerator and denominator by \(1+t^2\): \(p = \frac{2t}{(1+t^2) + (1-t^2) + 2t} = \frac{2t}{2 + 2t} = \frac{t}{1+t}\).

Simplify q

Using the same substitutions for \(\sin \theta\) and \(\cos \theta\) in the expression for \(q\):

\(q = \frac{1 + \frac{2t}{1+t^2}}{1 + \frac{2t}{1+t^2} - \frac{1-t^2}{1+t^2}}\)

Multiply the numerator and denominator by \(1+t^2\): \(q = \frac{(1+t^2) + 2t}{(1+t^2) + 2t - (1-t^2)} = \frac{(1+t)^2}{1+t^2+2t-1+t^2} = \frac{(1+t)^2}{2t+2t^2} = \frac{(1+t)^2}{2t(1+t)}\).

Simplifying further, we get: \(q = \frac{1+t}{2t}\).

p and q Reciprocals

We calculate the reciprocal terms required for the final expression:

• \(\frac{1}{p} = \frac{1}{\frac{t}{1+t}} = \frac{1+t}{t}\)
• \(\frac{1}{q} = \frac{1}{\frac{1+t}{2t}} = \frac{2t}{1+t}\)

Evaluate (p + 1/q)

Now, we find the value of the first bracket term \(p+\frac{1}{q}\):

\(p+\frac{1}{q} = \frac{t}{1+t} + \frac{2t}{1+t} = \frac{t+2t}{1+t} = \frac{3t}{1+t}\).

Evaluate (q + 1/p)

Next, we find the value of the second bracket term \(q+\frac{1}{p}\):

\(q+\frac{1}{p} = \frac{1+t}{2t} + \frac{1+t}{t} = \frac{1+t}{2t} + \frac{2(1+t)}{2t} = \frac{(1+t) + 2(1+t)}{2t} = \frac{3(1+t)}{2t}\).

Multiply Bracket Terms

Finally, we multiply the results of the two bracket terms:

\(\left(p+\frac{1}{q}\right)\left(q+\frac{1}{p}\right) = \left(\frac{3t}{1+t}\right) \left(\frac{3(1+t)}{2t}\right)\)

\(= \frac{3t \cdot 3(1+t)}{(1+t) \cdot 2t} = \frac{9t(1+t)}{2t(1+t)}\).

Final Value

By cancelling the common terms \(t\) and \((1+t)\) (assuming \(t \neq 0\) and \(t \neq -1\)), the expression simplifies to:

\(\frac{9}{2}\).