Question

If $\tan A=\sqrt{3}$, then verify that ${\sin }^{2}A+{\cos }^{2}A=1$.

Answer 1

$\tan A=\sqrt{3}$ (Given)

So $\angle A={60}^o$

Now ${\sin }^{2}A{\cos }^{2}A$

$={\sin }^2{60}^o{\cos }^2{60}^o$

$={\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2$

$=\frac{3}{4}+\frac{1}{4}$

$=\frac{4}{4}=1$

$\therefore {\sin }^{2}A+{\cos }^{2}A=1$

Hence Proved.