10.cos(A+B) + sin(A - B)=2 sin(45° + A) cos(45° + B)

1.	10.cos(A+B) + sin(A - B)=2 sin(45° + A) cos(45° + B)
Answer» Given Cos(A + B) + sin(A - B) = 2sin(45° + A)cos(45° + B) lift hand side ( LHS ) Cos(A + B) + sin(A - B) cosAcosB - sinAsinB + sinAcosB - COSASINB RHS => 2(sin45°cosA + cos45°sinA)(cos45cosB - sin45sinB) =>2(√2/2cosA + √2/2sinA)(√2/2cosB - √2/2sinB) => 2( (1/2)cosAcosB - (1/2)cosAsinB + (1/2)sinAcosB - (1/2)sinAsinB) => cosAcosB - cosAsinB + sinAcosB - sinAsinB LHS = RHS HOPE IT'S HELP

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10.cos(A+B) + sin(A - B)=2 sin(45° + A) cos(45° + B)​

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10.cos(A+B) + sin(A - B)=2 sin(45° + A) cos(45° + B)