Simplify (2sin ((π)/4 + θ) sin ((π)/4 - θ))/(cos 2θ)

\(\frac{2sin(\frac{π}{4} + θ) sin (\frac{π}{4} - θ)}{cos 2θ}\)

⇒ \(\frac{cos 2 - cos (\frac{π}{2})}{cos 2 θ}\)

( because 2sinAsinB = cos (A-B) - cos ( A+B))

⇒ \(\frac{cos2 θ - 0}{cos 2 θ}\) = 1