To find the value of \( \cot^2 \beta \), we begin with the given expressions:

• \( p \sin^2 \alpha + q \cos^2 \alpha = m \)
• \( p \cos^2 \beta + q \sin^2 \beta = n \)

We need to express \( \cot^2 \beta \) in terms of the given parameters.

The expression for \( \cot^2 \beta \) can be derived from the trigonometric identity:

• \( \cot \beta = \frac{\cos \beta}{\sin \beta} \)

Thus,

\(\cot^2 \beta = \frac{\cos^2 \beta}{\sin^2 \beta}\)

We substitute the second given equation into the expression for \( \cos^2 \beta \) and \( \sin^2 \beta \):

• From \( p \cos^2 \beta + q \sin^2 \beta = n \), we know:
  • \( \cos^2 \beta = \frac{n-q \sin^2 \beta}{p} \)

Using the identity \( \cos^2 \beta + \sin^2 \beta = 1 \), we express \( \cos^2 \beta \) in another form:

\(\cos^2 \beta = 1 - \sin^2 \beta\)

Thus, substituting this into the expression for \( \cot^2 \beta \):

• \( \cot^2 \beta = \frac{1 - \sin^2 \beta}{\sin^2 \beta} \)

Let \( \sin^2 \beta = x \); therefore, \( \cot^2 \beta = \frac{1-x}{x} \).

From \( p \cos^2 \beta + q \sin^2 \beta = n \), replace \( \cos^2 \beta \):

• \( p(1-x) + qx = n \)

This simplifies to:

\(p - px + qx = n\)

Rearranging it:

\(x(q - p) = n - p\)

So:

\(x = \frac{n-p}{q-p}\)

Substitute \( x \) back into the expression for \( \cot^2 \beta \):

• \( \cot^2 \beta = \frac{1-\frac{n-p}{q-p}}{\frac{n-p}{q-p}} = \frac{q-p-(n-p)}{n-p} = \frac{q-n}{n-p} \)

Thus, the required value of \( \cot^2 \beta \) is:

\(\frac{n-q}{n-p}\)

Therefore, the correct answer is \(\frac{n-q}{n-p}\).