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Question

In a right angled triangle ABC, write the value of sin2 A+sin2 B+sin2 C.

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Solution

Suppose in ΔABC B=90

A+C=π2

A=π2C

sin A=sin (π2C)=cos C Now, sin2 A+sin2 B+sin2C

=sin2A+1+cos2A[ sin π2=1]

= 1 + 1 = 2


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