92
If $\tan A = \sqrt{3}$; where $A$ is an acute angle, then find the value of $\frac{\sin^2 A}{1 + \cos^2 A}$.
Show SolutionHide Solution↓
$\tan A = \sqrt{3} = \tan 60^\circ$
$\Rightarrow A = 60^\circ$
$\frac{\sin^2 A}{1+\cos^2 A} = \frac{\sin^2 60^\circ}{1+\cos^2 60^\circ}$
$= \frac{(\frac{\sqrt{3}}{2})^2}{1+(\frac{1}{2})^2}$
$= \frac{\frac{3}{4}}{1+\frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}}$
$= \frac{3}{5}$
$\Rightarrow A = 60^\circ$
$\frac{\sin^2 A}{1+\cos^2 A} = \frac{\sin^2 60^\circ}{1+\cos^2 60^\circ}$
$= \frac{(\frac{\sqrt{3}}{2})^2}{1+(\frac{1}{2})^2}$
$= \frac{\frac{3}{4}}{1+\frac{1}{4}} = \frac{\frac{3}{4}}{\frac{5}{4}}$
$= \frac{3}{5}$