If $$\frac{\sin^2 \phi - 3 \sin \phi + 2}{\cos^2 \phi} = 1$$ where $$0^\circ < \phi < 90^\circ$$, then what is the value of $$(\cos 2 \phi + \sin 3 \phi + \cosec 2 \phi)$$?

$$\frac{\sin^2 \phi - 3 \sin \phi + 2}{\cos^2 \phi} = 1$$
On putting the $$\phi = 30\degree$$,
$$\frac{\sin^230\degree - 3 \sin 30\degree + 2}{\cos^2 30\degree} = 1$$
$$\frac{\frac{1}{4} - 3 \times \frac{1}{2} + 2}{ \frac{3}{4}} = 1$$
$$\frac{1 - 6 + 8}{3} = 1 $$
$$\frac{3}{3} = 1 $$
1 = 1
Now,
$$(\cos 2 \phi + \sin 3 \phi + \cosec 2 \phi)$$
On putting the $$\phi = 30\degree$$,
= $$(\cos 2 \times 30\degree + \sin 3 \times  30\degree + \cosec 2 \times 30\degree)$$
= $$(\cos 60\degree + \sin 90\degree + \cosec 60\degree)$$
= $$\frac{1}{2} + 1 + \frac{2}{\sqrt3}$$
= $$\frac{3\sqrt3 + 4}{2\sqrt3}$$
= $$\frac{3\sqrt3 + 4}{2\sqrt3} \times \frac{2\sqrt3}{2\sqrt3}$$
= $$\frac{18 + 8 \sqrt 3}{12}$$
= $$\frac{9 + 4 \sqrt 3}{6}$$