If $$2 \sin \theta + 15 \cos^2 \theta = 7, 0^\circ < \theta < 90^\circ$$, then $$\tan \theta + \cos \theta + \sec \theta =$$

$$2 \sin \theta + 15 \cos^2 \theta = 7$$

$$2 \sin \theta + 15 (1 - sin^2 \theta) = 7$$

$$2 \sin \theta + 15 - 15 sin^2 \theta = 7$$

$$2 \sin \theta + 8 - 15 sin^2 \theta = 0$$

$$15 sin^2 \theta - 2 \sin \theta - 8 = 0 $$

$$sin^2 \theta - \frac{2\sin \theta}{15} - \frac{8}{15} = 0 $$

$$sin^2 \theta + \frac{2\sin \theta}{3} - \frac{4\sin \theta}{5} - \frac{8}{15} = 0 $$

$$sin \theta(sin \theta + \frac{2}{3}) - \frac{4}{5}(sin \theta + \frac{2}{3}) = 0 $$

$$(sin \theta - \frac{4}{5})(sin \theta + \frac{2}{3}) = 0$$

For  $$0^\circ < \theta < 90^\circ$$,

$$sin \theta = \frac{4}{5}$$

Perpendicular = 4

Hypotenuses = 5

By triplet 3-4-5,

Base = 3 

Now,

$$\tan \theta + \cos \theta + \sec \theta$$

= $$\frac{4}{3} + \frac{3}{5} +\frac{5}{3}$$

= $$\frac{20 + 9 + 25}{15}$$

= $$\frac{54}{15}$$ = $$\frac{18}{5}$$

= $$3\frac{3}{5}$$