If sin x = $$\frac 35$$​ and x ∈ (0, $$\frac{\pi}2$$​), then find $$\frac{1+\tan x}{1-\tan x}$$​​

Correct option is B

Given: 
sin x = $$\frac 35$$​​ and x ∈ (0, $$\frac{\pi}2$$​)

Formula Used:
​$$\tan x = \frac{\sin x}{\cos x}\\\ \\\frac{1+\tan x}{1-\tan x}$$​​
Solution:
Since x lies in the first quadrant, all trigonometric ratios are positive.
​$$\sin x = \frac{3}{5} \\\cos x = \sqrt{1-\sin^2 x} = \sqrt{1-\left(\frac{3}{5}\right)^2}= \sqrt{\frac{16}{25}}= \frac{4}{5} \\\tan x = \frac{3/5}{4/5}= \frac{3}{4} \\\frac{1+\tan x}{1-\tan x}= \frac{1+\frac{3}{4}}{1-\frac{3}{4}}= \frac{7/4}{1/4}= 7$$​​
Therefore, the required value is 7.