Trigonometry – II — Maths STD 11 Science — Question
Maharashtra BoardEnglish MediumSTD 11 ScienceMathsTrigonometry – II2 Marks
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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