Question

State True/False

In any $\Delta ABC$ where $\angle B={90}^{\circ },si{n}^{2}A+co{s}^{2}C=1$

• A. True
• B. False

Answer 1

The correct option is B

False

Given that, $si{n}^{2}A+co{s}^{2}C=1$. But we know that $si{n}^{2}C+co{s}^{2}C=1$

$\therefore$ if $si{n}^{2}A+{\cos }^{2}C=1$, then $\angle A=\angle C$. But $\angle A=\angle C$ only if $△ABC$ is isosceles.
Therefore the above statement is not always true.