Question

prove that Sin(90° -A) = COSA​

Answer 1

We already know that,

Sin(P-Q) = SinP CosQ - SinQ CosP

so, using this trignometric identity we can write,

Sin(90°-A) = Sin90° CosA - SinA Cos90° ____1

we know, value of Sin90° = 1

Value of cos90° = 0

Putting the values in equation (1) We get :-

Sin(90°-A) = 1 CosA - SinA×0

Sin(90°-A) = CosA - 0

Sin(90°-A) = CosA

Hence proved that Sin(90°-A) = CosA

Hope it helps.