Question

In a $\Delta ABC,$ if $\angle C={90}^{\circ },$ prove that ${\sin }^{2}A+{\sin }^{2}B=1$

Answer 1

In $\Delta ABC$

$\angle C={90}^{\circ }$

So by Angle sum property $A+B={90}^{\circ }B=90-A$

${\sin }^{2}A+{\sin }^{2}B{\sin }^{2}A+{\sin }^2(90-A){\sin }^{2}A+{\cos }^{2}A=1$

Hence proved