(a) Start with the left-hand side (LHS):

\(\frac{\tan \theta + 7}{\tan^2 \theta - 3}\)

Use \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and \(\tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta}\):

\(\frac{\frac{\sin \theta}{\cos \theta} + 7}{\frac{\sin^2 \theta}{\cos^2 \theta} - 3} = \frac{\sin \theta + 7 \cos \theta}{\sin^2 \theta - 3 \cos^2 \theta}\)

Use \(\sin^2 \theta + \cos^2 \theta = 1\) to rewrite the denominator:

\(\sin^2 \theta - 3 \cos^2 \theta = (1 - \cos^2 \theta) - 3 \cos^2 \theta = 1 - 4 \cos^2 \theta\)

Thus, the LHS becomes:

\(\frac{\sin \theta \cos \theta + 7 \cos^2 \theta}{1 - 4 \cos^2 \theta}\)

This matches the right-hand side (RHS), proving the identity.

(b) Using the identity from part (a), solve:

\(\frac{\tan \theta + 7}{\tan^2 \theta - 3} = \frac{5}{\tan \theta}\)

Cross-multiply to eliminate fractions:

\(\tan \theta (\tan \theta + 7) = 5 (\tan^2 \theta - 3)\)

\(\tan^2 \theta + 7 \tan \theta = 5 \tan^2 \theta - 15\)

Rearrange to form a quadratic equation:

\(4 \tan^2 \theta - 7 \tan \theta - 15 = 0\)

Factorize:

\((4 \tan \theta + 5)(\tan \theta - 3) = 0\)

Solutions for \(\tan \theta\):

\(\tan \theta = -1.25 \quad \text{and} \quad \tan \theta = 3\)

Find \(\theta\) in the range \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 71.6^\circ\) and \(\theta = 128.7^\circ\).