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

Correct option is A

Given: 

$$tan\theta = \frac{8}{15}$$​,

Solution: 

​$$tan\theta = \frac{8}{15} = \frac{AB}{BC}$$ 

By Pythagoras Theorem:

​$$AC = \sqrt{(AB^2 + BC^2)}\\AC = \sqrt{(8^2 + 15^2 )}\\AC = \sqrt{(64+225)}\\AC = \sqrt{289}\\AC = 17$$ 

Then the value of $$\frac{\sqrt{1 - sin\theta}}{\sqrt{1 + sin\theta}}$$​ 

=  $$\frac{\sqrt{ (1- \frac{8}{17}})}{\sqrt {(1+ \frac{8}{17})}}$$ 

= $$\sqrt{\frac{\frac{17- 8}{17}}{\frac {17 +8}{17}}}$$

=   $$\sqrt{\frac{9}{25}}$$ 

= $$\frac{3}{5}$$​​