We have, L.H.S. = cos $\left(\frac{3\pi}{4}+x\right)$ - cos $\left(\frac{3\pi}{4}-x\right)$ = 2 sin $\left[\frac{\frac{3\pi}{4}+x+\frac{3\pi}{4}-x}{2}\right]$ sin $\left[\frac{\left(\frac{3\pi}{4}-x\right)-\left(\frac{3\pi}{4}+x\right)}{2}\right]$ Since cos A - cos B = 2 sin $\left(\frac{A+B}{2}\right)$ sin $\left(\frac{B-A}{2}\right)$ = 2 sin $\left(\frac{3\pi}{2} \times \frac{1}{2}\right)$ sin (- x) = 2 sin $\left(\frac{3\pi}{4}\right)$ sin (- x) = - 2 sin $\left(\frac{4\pi-\pi}{4}\right)$ sin x (Since sin (- x) = - sin x) = - 2 sin $\left(\pi - \frac{\pi}{4}\right)$ sin x = - 2 sin $-\frac{2}{\sqrt{2}}$ sin x [Since $\sqrt{2}$ lies in second quadrant in which sin x is +ve = $-2/√2$ sin x = - $√2$ sin x = R.H.S.