\(2sin\left(\frac{\pi}{4}+a\right)sin\left(\frac{\pi}{4}-a\right)=cos2a-cos\left(\frac{\pi}{2}\right)=cos2a\)
\(tanx-\frac{1}{tanx}=\frac{sinx}{cosx}-\frac{cosx}{sinx}=\frac{sin^2x-cos^2x}{sinx.cosx}=-\frac{2\left(cos^2x-sin^2x\right)}{2sinx.cosx}=\frac{2cos2x}{sin2x}=-2cot2x=-\frac{2}{tan2x}\)