Question

In $\Delta ABC,\angle C={90}^{\circ }$, then the value of ${\cos }^{2}A+{\cos }^{2}B$ is :

• A. $1$
• B. $0$
• C. $\frac{1}{2}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is A $1$

As $\angle C={90}^o$

$\therefore cosA=\frac{BC}{AB}$

$cosB=\frac{AC}{AB}$

${cos}^{2}A+{cos}^{2}B=\frac{{BC}^2}{{AB}^2}+\frac{{AC}^2}{{AB}^2}=\frac{{BC}^2+{AC}^2}{{AB}^2}$

From $△ABC$,

By Pythagoras Theorem,

${AB}^2={AC}^2+{BC}^2$

$\Rightarrow {cos}^{2}A+{cos}^{2}B=\frac{{AB}^2}{{AB}^2}$

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