If $$tan θ= \frac{12}5$$​, then $$\frac{(1+sinθ)}{(1- sin θ )}$$​is:

Correct option is C

GIVEN:
​$$tan θ= \frac{12}5,$$​​
Solution:
Here using A instead of  $$tan θ= \frac{12}5,$$​​
Cosec² A = 1 + cot² A
= 1 + ( $$\frac{5}{12}$$​ )²

= 1 + $$\frac{25}{144}$$​= $$\frac{ 144 + 25 } { 144}$$​​
= $$\frac{169} { 144}$$​​

cosecA = √ ( $$\frac{13}{12}$$​ )²
​$$\frac{1}{sinA} = \frac{13}{12}$$​​
Therefore ,
SinA = $$\frac{12}{13}$$​​
​$$\frac{1+sin A}{1-sinA}=\frac{1+\frac{12}{13}}{1-\frac{12}{13}}$$​​
= $$\frac{\frac{25}{13}}{\frac{1}{13}}$$​​
=25