Question

In a $△ABC,\angle A=\frac{\pi }{2},$ then ${\cos }^{2}B+{\cos }^{2}C$ equals :

• A. $-2$
• B. $1$
• C. $-1$
• D. $zero$

Answer 1

Answer: (B)

$1$

We have,

$In\Delta ABC$ $\angle A=\frac{\pi }{2}$

We know that,

$In\Delta ABC$

$\angle A+\angle B+\angle C=\pi$

$\angle B+\angle C=\pi -\frac{\pi }{2}$

$\angle B+\angle C=\frac{\pi }{2}$

$\angle B=\frac{\pi }{2}-\angle C$

Now,

${\cos }^{2}B+{\cos }^{2}C=?$

${\cos }^{2}B+{\cos }^{2}C$

$={\cos }^2(\frac{\pi }{2}-C)+{\cos }^{2}C$

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

$=1$

Hence, this is the answer.