1.

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

Answer»

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 sin2A + sin2B + sin2C

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 = (sin 2A + cos 2A) + 1

= 1 + 1

= 2

[by using the identity sin2θ + cos2θ = 1]

Therefore, sin2A + sin2B + sin2C = 2.



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