If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and
$\mathrm{M}=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$, then :
Correct Option: 1
$L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$
$\left(\because \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\right)$
$\Rightarrow \mathrm{L}=\left(\frac{1-\cos (\pi / 8)}{2}\right)-\left(\frac{1-\cos (\pi / 4)}{2}\right)$
$\mathrm{L}=\frac{1}{2}\left[\cos \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{8}\right)\right]$
$\mathrm{L}=\frac{1}{2 \sqrt{2}}-\frac{1}{2} \cos \left(\frac{\pi}{8}\right)$
$\mathrm{M}=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$
$M=\frac{1+\cos (\pi / 8)}{2}-\frac{1-\cos (\pi / 4)}{2}$
$\mathrm{M}=\frac{1}{2} \cos \left(\frac{\pi}{8}\right)+\frac{1}{2 \sqrt{2}}$