Answer: (a) identity proved. (b) \(9y^2-x^2y^2=81\).

(a) Start with the left-hand side:

\(\frac{\tan^2\theta+\sin^2\theta}{\cos\theta+\operatorname{sec}\theta}.\)

Write everything in terms of sine and cosine:

\(\tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}, \qquad \operatorname{sec}\theta=\frac1{\cos\theta}.\)

Then

\(\frac{\tan^2\theta+\sin^2\theta}{\cos\theta+\operatorname{sec}\theta} = \frac{\frac{\sin^2\theta}{\cos^2\theta}+\sin^2\theta}{\cos\theta+\frac1{\cos\theta}}.\)

Simplify numerator and denominator:

\(\frac{\sin^2\theta\left(\frac{1+\cos^2\theta}{\cos^2\theta}\right)}{\frac{\cos^2\theta+1}{\cos\theta}} = \sin^2\theta\cdot\frac{1+\cos^2\theta}{\cos^2\theta}\cdot\frac{\cos\theta}{1+\cos^2\theta}.\)

Cancel \(1+\cos^2\theta\):

\(\frac{\sin^2\theta}{\cos\theta} = \sin\theta\cdot\frac{\sin\theta}{\cos\theta} = \tan\theta\sin\theta.\)

So the identity is proved.

(b) Since \(x=3\sin\phi\), we have

\(x^2=9\sin^2\phi.\)

Since \(y=\frac3{\cos\phi}\), we have

\(y^2=\frac9{\cos^2\phi}.\)

Therefore

\(9y^2-x^2y^2 = y^2(9-x^2) = \frac9{\cos^2\phi}\left(9-9\sin^2\phi\right).\)

Factor out \(9\):

\(9y^2-x^2y^2 = \frac{81(1-\sin^2\phi)}{\cos^2\phi}.\)

Using \(1-\sin^2\phi=\cos^2\phi\),

\(9y^2-x^2y^2=81.\)