Cosπ4+A.Cosπ4-B-Sinπ4+A.Sinπ4-B=

Cosπ4+A Cosπ4-B-Sinπ4+A Sinπ4-B =cosπ4+A+π4-B =cosπ2+A-B =-sin(A-B) =sin(B-A)

Cosπ4+A.Cosπ4-B-Sinπ4+A.Sinπ4-B=

Cosπ4+A Cosπ4-B-Sinπ4+A Sinπ4-B =cosπ4+A+π4-B =cosπ2+A-B =-sin(A-B) =sin(B-A)

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$Cos (\frac{π}{4} + A) . Cos (\frac{π}{4} - B) - Sin (\frac{π}{4} + A) . Sin (\frac{π}{4} - B) =$

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$Cos (A + B)$

$Cos (A - B)$

$Sin (A + B)$

$Sin (B - A)$

answer is D.

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Detailed Solution

$\begin{array}{rcl} Cos (\frac{π}{4} + A) Cos (\frac{π}{4} - B) - Sin (\frac{π}{4} + A) Sin (\frac{π}{4} - B) \\ = & \cos (\frac{π}{4} + A + \frac{π}{4} - B) \\ = & \cos (\frac{π}{2} + A - B) \\ = & - \sin (A - B) \\ = & \sin (B - A) \end{array}$