Question

In $△ABC,\angle C={90}^{\circ }$, then the value of $si{n}^{2}A+si{n}^{2}B$ is ?

• A. 1
• B. $\frac{1}{4}$
• C. $\frac{3}{4}$
• D. 0

Answer 1

Answer: (A)

1

In $△ABC$

$\angle A+\angle B+\angle C={180}^{\circ }\angle A+\angle B+{90}^{\circ }={180}^{\circ }\angle B={90}^{\circ }-\angle A$

${\sin }^{2}A+{\sin }^{2}B={\sin }^{2}A+{\sin }^2({90}^{\circ }-\angle A)$

Now $\sin ({90}^{\circ }-\angle A)=\cos A$

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