If 5 sinθ = 3, the numerical value of $$\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}$$ is

Expression : $$5 sin\theta = 3$$

=> $$sin\theta = \frac{3}{5}$$

We know that, $$cos\theta = \sqrt{1 - sin^2\theta}$$

=> $$cos\theta = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}}$$

=> $$cos\theta = \frac{4}{5}$$

Now, $$tan\theta = \frac{3}{4}$$ and $$sec\theta = \frac{5}{4}$$

To find : $$\frac{\sec\theta-\tan\theta}{\sec\theta+\tan\theta}$$

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

= $$\frac{2}{8} = \frac{1}{4}$$