If $$5 \sin^2 \theta + 14 \cos \theta = 13, 0^\circ < \theta < 90^\circ$$, then what is the value of $$\frac{\sec \theta + \cot \theta}{\cosec \theta + \tan \theta}$$?

$$5 \sin^2 \theta + 14 \cos \theta = 13$$

$$5(1 - cos^2 \theta)+ 14 \cos \theta = 13$$

$$-5cos^2 \theta + 14 \cos \theta = 8$$

$$5cos^2 \theta - 10 \cos \theta - 4 \cos \theta + 8 = 0$$

$$5cos \theta(\cos \theta - 2)  - 4 (\cos \theta - 2) = 0$$

$$(5cos \theta - 4)(\cos \theta - 2) = 0$$

$$(\cos \theta - 2) $$(it is not possible)

For $$0^\circ < \theta < 90^\circ$$,

$$(5cos \theta - 4) = 0$$

$$cos \theta = 4/5$$

by triplet 3-4-5,

$$\sin \theta = 3/5$$

Now,

$$\dfrac{\sec \theta + \cot \theta}{\cosec \theta + \tan \theta}$$

= $$\dfrac{\sec \theta(1 + \frac{1}{\sin \theta})}{\sec \theta(\cosec \theta\cos \theta + \sin \theta)}$$

= $$\dfrac{1 + \frac{1}{\sin \theta}}{\cosec \theta\cos \theta + \sin \theta}$$

= $$\dfrac{1 + \frac{5}{3}}{\frac{5}{3} \times \frac{4}{5} + \frac{3}{5}}$$

= $$\dfrac{\frac{8}{3}}{ \frac{4}{3} + \frac{3}{5}}$$
= $$\dfrac{\frac{8}{3}}{\frac{29}{15}}$$ = $$\dfrac{40}{29}$$