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Question

If cosA+sinB=m and sinA+cosB=n, prove that 2sin(A+B)=m2+n22.

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Solution

Given: cosA+sinB=m and sinA+cosB=n
Now,
2sin(A+B)
=2(sinAcosB+cosAsinB)
=2+2sinAcosB+2cosAsinB2
=sin2A+cos2A+sin2B+cos2B
+2sinAcosB+2cosAsinB2
=sin2A+cos2B+2sinAcosB
+cos2A+sin2B+2cosAsinB2
=(sinA+cosB)2+(cosA+sinB)22
=n2+m22
Hence proved

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