If $$tan\theta=\frac{8}{15}$$, the value of $$\sqrt{\frac{1-sin\theta}{1+sin\theta}}$$ is

Given : $$tan\theta=\frac{8}{15}$$

Using, $$sec^2\theta-tan^2\theta=1$$

=> $$sec^2\theta=1+(\frac{8}{15})^2=1+\frac{64}{225}$$

=> $$sec^2\theta=\frac{225+64}{225}=\frac{289}{225}$$

=> $$sec\theta=\sqrt{\frac{289}{225}}=\frac{17}{15}$$

To find : $$\sqrt{\frac{1-sin\theta}{1+sin\theta}}$$

= $$\sqrt{\frac{1-sin\theta}{1+sin\theta}} \times \sqrt{\frac{1-sin\theta}{1-sin\theta}}$$

= $$\sqrt{\frac{(1-sin\theta)^2}{(1-sin\theta)(1+sin\theta)}}=\sqrt{\frac{(1-sin\theta)^2}{1-sin^2\theta}}$$

= $$\sqrt{\frac{(1-sin\theta)^2}{cos^2\theta}}=\frac{1-sin\theta}{cos\theta}$$

= $$\frac{1}{cos\theta}-\frac{sin\theta}{cos\theta}=sec\theta-tan\theta$$

= $$\frac{17}{15}-\frac{8}{15}=\frac{17-8}{15}$$

= $$\frac{9}{15}=\frac{3}{5}$$

=> Ans - (C)