In a right angled triangle ABC, write the value of sin2 A+sin2 B+sin2 C.
Suppose in ΔABC ∠B=90∘
⇒ A+C=π2
⇒ A=π2−C
⇒ sin A=sin (π2−C)=cos C Now, sin2 A+sin2 B+sin2C
=sin2A+1+cos2A[∵ sin π2=1]
= 1 + 1 = 2