Question

if cos A +sin B =m and sin A +cos B=n,prove that 2 sin (A+B)=mxm +nxn-2

Answer 1

given ,sinA=+cosB=m
and cosA+sinB=n
to prove;sin(A+B)=mn+n×n-2
we have to find first the value of m×n
m×n=(sinA+cosB)(cosA+sinB)=sinAcosA+sinACosB +cosAcosB+cosBsinB
=(cosAcosB+sinAsinB)+(sinAcosB+cosB sinA)
=cos (A-B)+sin(A+B)
now we have to find
n×n=(cosA+sinB)²= cos²A+sin²B+2sinB.cosA
put the value in RHS,
m×n+n×n-2=sin(A+B)+cos(A-B)+cos²A+sin²B+2sinB cosA-2
=sin(A+B)+ cos(A-B)+cos²A +sin²B + 2sinBcosA-1-1
= sin(A+B)+cos(A-B)+sin²B+cos²A+2sinB.cosA-sin²A-cos²A-sin²B-cos²B
=sin(A+B)+cos(A-B)+2sinB.cosA+sin²B-sin²B+cos²A-cos²A-sin²A-cos²B
=sin(A+B)+cos(A-B)+2sinBcosA-(sin²A+cos²B)
=sin(A+B)+cos(A-B)+sin(A+B)+sin(A-B)-