MCQ
In $\triangle ABC , \angle A =\frac{\pi}{2}$, then $\cos ^2 B+\cos ^2 C =$
  • A
    -2
  • B
    -1
  • C
    $0$
  • 1

Answer

Correct option: D.
1
(D)
Given, $\angle A =\frac{\pi}{2}$
In $\triangle ABC , \angle A +\angle B +\angle C =\pi$
$\therefore\angle B +\angle C =\frac{\pi}{2}$
$\Rightarrow B =\frac{\pi}{2}- C$
$\Rightarrow \cos ^2 B=\cos ^2\left(\frac{\pi}{2}- C \right)=\sin ^2 C$
$\therefore \cos ^2 B+\cos ^2 C =\sin ^2 C +\cos ^2 C =1$

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