If sin alpha= cos beta, then prove that alpha+beta=90

     sin α = cos βsin α = sin90°-βα = 90° - βα + β = 90°

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Sin alpha = cos beta (given), We know that cos theta =sin (90-theta). I.e., sin alpha = sin (90- beta) -> alpha = 90- beta -> alpha + beta =90
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Given : sin A = cos B
so, (sinA)^2 = (cosB)^2
=》sin^2 A = cos^2 B
=》1-cos^2 A = 1-sin^2 B
=》-cos^2 A = -sin^2 B
=》cos^2 A = sin^2 B
=》cos A = sin B
so, sin A = cos B and sin B = cos A
Now,
sin(A+B) = sinA×cosB + sin B × cos A
= sinA×sinA + cosA× cosA [sin A = cos B]
=》sin(A+B) = sin^2 A + cos^2 A = 1
=》sin(A+B) = 1
Also, sin 90° = 1
so, sin(A+B) = sin 90° = 1
=》sin(A+B) = sin90
=》A+B = 90°
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