\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\f

1.	\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\frac{\pi}{4}-y\right)-} \\ {\sin \left(\frac{\pi}{4}-x\right) \cdot \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)}\end{array}
Answer» LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y) Let ( π/4 -x) = A (π/4 -y) = BThen, LHS = cosA.cosB -sinA.sinB But we Know, cos(A + B) = cosA.cosB-sinA.sinB use this, = cos(A + B) = cos{(π/4 -x)+(π/4 -y)}=cos(π/2 -(x +y)}We know, Cos(π/2 -∅) = sin∅ use this , = sin(x + y) = RHS

\begin{array}{l}{\cos \left(\frac{\pi}{4}-x\right) \cdot \cos \left(\frac{\pi}{4}-y\right)-} \\ {\sin \left(\frac{\pi}{4}-x\right) \cdot \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)}\end{array}

Answer»

LHS = cos(π/4 - x).cos(π/4-y)-sin(π/4 -x).sin(π/4-y)

Let ( π/4 -x) = A (π/4 -y) = BThen, LHS = cosA.cosB -sinA.sinB But we Know, cos(A + B) = cosA.cosB-sinA.sinB use this,

= cos(A + B) = cos{(π/4 -x)+(π/4 -y)}=cos(π/2 -(x +y)}We know, Cos(π/2 -∅) = sin∅ use this , = sin(x + y) = RHS