If sinA = $$\frac{4}{\sqrt{17}}$$​ then tan 2 A = ? (0° < A < 90°)

Correct option is A

Given:
​$$\sin A = \frac{4}{\sqrt{17}} \quad \text{and} \quad 0^\circ < A < 90^\circ$$​
Concept Used:
​$$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$$

$$\sin^2 A + \cos^2 A = 1$$​​
Solution:
​$$\sin^2 A + \cos^2 A = 1$$​
​$$\left(\frac{4}{\sqrt{17}}\right)^2 + \cos^2 A = 1$$​
​$$\frac{16}{17} + \cos^2 A = 1$$​
​$$\cos^2 A = 1 - \frac{16}{17} = \frac{1}{17}$$​
​$$\cos A = \frac{1}{\sqrt{17}}$$​
​$$\tan A = \frac{\sin A}{\cos A} = \frac{\frac{4}{\sqrt{17}}}{\frac{1}{\sqrt{17}}} = 4$$​
​$$\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}$$

​​$$\tan 2A = \frac{2 \times 4}{1 - 4^2} = \frac{8}{1 - 16} = \frac{8}{-15} = -\frac{8}{15}$$​