Question

If cos (90°-A)=0,then sin (A+B)=​

Answer 1

Answer:

Given: Cos (90°-A)= 0

To find: Sin(A+B)

Solution:

Cos(90°-A)= 0

→ Sin A= 0

→A= nπ

sin(A+B)=sin(nπ+B)= +sin B

Therefore, Sin(A+B)= + SinB

Answer 2

Given,

cos(90°-A) = 0

sinA = 0 { cos(90°-A) = sinA}

To find; sin(A+B) =?

According to question,

sin(A+B)

= sinA.cosB + cosA.sinB

= 0.cosB + cosA.sinB

= 0 + cosA.sinB

= cosA.sinB

Therefore, sin(A+B) = cosA.sinB