If sin A = \frac{3}{5} and A lies in the quadrant, what is the value of Tan A?

Given:
$\sin A = \frac{3}{5}$ and A is in second quadrant

Using identity:
$\sin^2 A + \cos^2 A = 1$

$\cos^2 A = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}$

$\cos A = \frac{4}{5}$

But in second quadrant, $\cos A$ is negative

So,
$\cos A = -\frac{4}{5}$

Now,
$\tan A = \frac{\sin A}{\cos A}$

$= \frac{\frac{3}{5}}{-\frac{4}{5}} = -\frac{3}{4}$

Final Answer: $-\frac{3}{4}$ ✅