Question : ABC is a right-angled triangle with $\angle A=90°$. Then the value of $\cos^{2}A+\cos^{2}B+cos^{2}C$ is:
Option 1: 2
Option 2: 1
Option 3: 0
Option 4: 3
Correct Answer: 1
Solution :
Given: $\angle A=90°$
We know that, $\angle A+\angle B+\angle C=180°$
Putting the value of $\angle A$
⇒ $90°+\angle B+\angle C=180°$
⇒ $\angle B+\angle C=180°–90°$
⇒ $\angle B+\angle C=90°$
⇒ $\angle B=90°–\angle C$
Now, $\cos^{2}A+\cos^{2}B+\cos^{2}C$
Putting the values, we get
$\cos^{2}90°+\cos^{2}(90°–\angle C)+\cos^{2}C$
$=0+\sin^{2} C+\cos^{2}C$
$=\sin^{2} C+\cos^{2}C = 1$
Hence, the correct answer is $1$.
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