23. If A + B = 90°, then sin²A + sin’B = A) 1 B) 0. C) 2 D)

1.	23. If A + B = 90°, then sin²A + sin’B = A) 1 B) 0. C) 2 D)
Answer» Answer: 1 Step-by-step EXPLANATION: GIVEN ∠C=90o From TRIANGLE, sinA=ABBC; sinB=ABAC ⇒sin2A+sin2B AB2BC2+AB2AC2=AB2BC2+AC2 From △ABC, by using Pythagoras THEOREM, AB2=BC2+AC2 sin2A+sin2B=AB2BC2+AC2=AB2AB2=1

Answer»

Answer:

Step-by-step EXPLANATION:

GIVEN ∠C=90o

From TRIANGLE,

sinA=ABBC; sinB=ABAC

⇒sin2A+sin2B

AB2BC2+AB2AC2=AB2BC2+AC2

From △ABC, by using Pythagoras THEOREM,

AB2=BC2+AC2

sin2A+sin2B=AB2BC2+AC2=AB2AB2=1

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23. If A + B = 90°, then sin²A + sin’B = A) 1 B) 0. C) 2 D)​

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23. If A + B = 90°, then sin²A + sin’B = A) 1 B) 0. C) 2 D)