Question

HEY LISTEN
:-1) if A and B are two acute angles,then prove that sin(A+B)= sinAcosB+cosAsinB
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Answer 1

given AandB are acute angles
let A=60,B=30
sin(60+30)=sin60cos30+cos60sin30
sin90 =√3/2.√3/2+1/2.1/2
1. =3/4+1/4
1. = 1
therefore we can that
sin(A+B)=sinAcosB+cosAsinB

Answer 2

In a right triangle one angle is 90°
Therefore, A+B=180°-90°=90°
Therefore,
LHS-
Sin(A+B)
=sin90°
=1
RHs-
sinAcosB+cosAsinB
or,sinAsin(90-B)+sin(90-A)sinB. (as,sinA=cos(90-A) and cosA=sin(90-A)
or,sinAsinA+sinBsinB
or,2sinA+2sinB
or,2sin(A+B)
or, 2sin90° [as,A+B=90°]
Or,2×1=1
=>LHS=RHS
Hence proved