if A and B are acute angles and sinA=cosB,prove that A+B=90degrees
plz help me
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RonL
Hello, sanya.batra!
Here's a roundabout approach . . .
Of A and B are acute angles and sin A = cos B,
prove that: A + B = 90°
We are given: .cos B .= .sin A . [1]
Square both sides: .cos²B .= .sin²A
Subtract from 1: .1 - cos²B .= .1 - sin²A
And we have: .sin²B .= .cos²A
Since A and B are acute: .sin B = cos A . [2]
Now consider: .sin(A + B)
. . . . . . . . . We have: .sin(A + B) .= .sin(A)·cos(B) + sin(B)·cos(A)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .↓ . . . . . ↓
Substitute [1] and [2]: .sin(A + B) .= .sinA·sin(A) + cos(A)·cos(A)
Hence: .sin(A + B) .= .sin²A + cos²A . → . sin(A + B) .= .1
Therefore: .A + B .= .90°
(Recall that A and B are both acute angles.)
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RonL