Question

In any $\Delta$ABC where $\angle$ B=90${}^{\circ }$, ${\sin }^{2}A$ + ${\cos }^{2}C$ = 1.

• A. True
• B. False

Answer 1

The correct option is B

False

As per question
In a $\Delta$ABC , $\angle B={90}^{\circ }$

Now,$\angle A+\angle C={90}^{\circ }$
So,$\angle C={90}^{\circ }-\angle A$
${\sin }^{2}A+{\cos }^{2}C={\sin }^{2}A+{\cos }^2({90}^{\circ }-A)={\sin }^{2}A+{\sin }^{2}A(\because cos({90}^{\circ }-\theta )=sin\theta )=2{\sin }^{2}A$
Now, ${\sin }^{2}A+{\cos }^{2}C=2{\sin }^{2}A=1onlywhen\angle A=\angle C={45}^{\circ }$
So it is not always true.