Question

$\Delta$ ABC is right angled at C. Then prove that ${\sin }^{2}A+{\sin }^{2}B=1$.

Answer 1

Given, $\Delta$ ABC is right angled at C.

Now $A+B+C={180}^o$

or, $A+B={90}^o$ [ Since $C={90}^o$]

or, $B={90}^o-A$......(1).

Now, ${\sin }^{2}A+{\sin }^{2}B$

$={\sin }^{2}A+{\sin }^2({90}^o-A)$

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

$=1$.