If $$\sin x = \frac{12}{37}$$, then what is the value of $$\tan x$$?

Given,   $$\sin x=\frac{12}{37}$$

$$\therefore\ \tan x=\frac{\sin x}{\cos x}$$

$$=\frac{\sin x}{\sqrt{1-\sin^2x}}$$

$$=\frac{\frac{12}{37}}{\sqrt{1-\left(\frac{12}{37}\right)^2}}$$

$$=\frac{12}{37\sqrt{1-\frac{144}{1369}}}$$

$$=\frac{12}{37\sqrt{\frac{1369-144}{1369}}}$$

$$=\frac{12}{37\sqrt{\frac{1225}{1369}}}$$

$$=\frac{12}{37\times\frac{35}{37}}$$

$$=\frac{12}{35}$$

Hence, the correct answer is Option D