Question

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

• A. -2
• B. 1
• C. 2
• D. Zero

Answer 1

The correct option is B

1

$co{s}^{2}B+co{s}^{2}C=co{s}^B+co{s}^2[\frac{\pi }{2}-B]$
$=co{s}^{2}B+si{n}^{2}B=1$