In a Right Angled Triangle ABC, Find the Value of sin²A + sin²B + sin²C
Question
In a right angled triangle ABC, find the value of
\[ \sin^2A+\sin^2B+\sin^2C. \]Solution
Since ABC is a right angled triangle,
\[ C=\frac{\pi}{2} \]Therefore,
\[ \sin^2C = \sin^2\frac{\pi}{2} = 1 \]Also,
\[ A+B=\frac{\pi}{2} \]Hence,
\[ \sin^2B = \cos^2A \]Substituting,
\[ \sin^2A+\sin^2B+\sin^2C = \sin^2A+\cos^2A+1 \] \[ =1+1 \] \[ =2 \]