Q. 7

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

Given, triangle ABC is right angle.

So, let B = 90°

Then as per the property of angles in a triangle

A + B + C = 180°

As B = 90°

A + 90° + C = 180°

Then A + C = 180° - 90° = 90°

Now, consider sin 2A + sin 2B + sin 2C

As B = 90°

sin2A + sin2B + sin2C = sin2A + sin2(90°) + sin2C

= sin2A + 1 + sin2C

From before, we know that A + C = 90° ; C = 90° - A

sin2A + sin2B + sin2C = sin2A + 1 + sin2( 90° - A)

= sin2A + cos2(A) + 1

[by using the identity cos x = sin ( 90° - x)]

sin2A + sin2B + sin2C = (sin2A + cos2A) + 1

= 1 + 1

= 2

[by using the identity sin2θ + cos2θ = 1]

Therefore, sin2A + sin2B + sin2C = 2.

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