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In a right-angled ∆ABC, sin^2A + sin^2B + sin^2C is equal to

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in Trigonometry by (53.3k points)

In a right-angled ∆ABC, sin2A + sin2B + sin2C is equal to

(A)  0 

(B)  1 

(C)  –1 

(D)  None of these 

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1 Answer

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Correct option (D) None of these

Let C = 90°, then

sin2A + sin2B + sin2C = sin2A + sin2B + 1

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