If A + B = 90° and sin A = $$\frac{3}{5}$$​, then the value of tan B is:

Correct option is D

Given:

A + B $$= 90^\circ$$​​

​$$\sin A = \frac{3}{5}$$​​

Formula Used:

​$$\sin^2 A + \cos^2 A = 1$$​​

​$$\tan B = \frac{\sin B}{\cos B}$$​​

Solution:

​$$\left( \frac{3}{5} \right)^2 + \cos^2 A = 1$$​

​$$\frac{9}{25} + \cos^2 A = 1$$​

​$$\cos^2 A = 1 - \frac{9}{25} = \frac{16}{25}$$​

​$$\cos A = \frac{4}{5}$$​

Since $$\sin B = \cos A$$​ and $$\cos B = \sin A$$​, we have:

​$$\sin B = \frac{4}{5}, \quad \cos B = \frac{3}{5}$$​​

​$$\tan B = \frac{\sin B}{\cos B} = \frac{\frac{4}{5}}{\frac{3}{5}} = \frac{4}{3}$$

Alternate Method:

​$$\sin A = \frac{3}{5}$$ =$$\frac{\text{Perpendicular}}{\text{Hypotaneous}}$$

(Hypotaneous)² = (Perpendicular)² + (Base)²​

(Base)² = $$5^2 -3^2 = 25-9 =16$$

Base = 4

tan B =$$\frac{\text{Perpendicular}}{\text{Base}}=\frac 43$$​​