If $L=\sin ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$ and $M=\cos ^{2}\left(\frac{\pi}{16}\right)-\sin ^{2}\left(\frac{\pi}{8}\right)$, then:

(C) $M=\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$

(A) $\mathrm{M}=\frac{1}{4 \sqrt{2}}+\frac{1}{4} \cos \frac{\pi}{8}$

(B) $L=-\frac{1}{2 \sqrt{2}}+\frac{1}{2} \cos \frac{\pi}{8}$

(D) $\mathrm{L}=\frac{1}{4 \sqrt{2}}-\frac{1}{4} \cos \frac{\pi}{8}$

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Trigonometric Ratios and Identities

L = sin^{2} (\frac{π}{16}) - sin^{2} (\frac{π}{8})

and

M = cos^{2} (\frac{π}{16}) - sin^{2} (\frac{π}{8})

, then:

Answer: (C)

Write

cos^{2} \frac{π}{16} = \frac{1 + c o s \frac{π}{8}}{2}

and

sin^{2} \frac{π}{8} = \frac{1 - c o s \frac{π}{4}}{2} = \frac{1 - \frac{1}{2}}{2}

. Then

M = cos^{2} \frac{π}{16} - sin^{2} \frac{π}{8} = \frac{1 + c o s \frac{π}{8}}{2} - \frac{1 - \frac{1}{2}}{2} = \frac{c o s \frac{π}{8} + \frac{1}{2}}{2} = \frac{1}{2 2} + \frac{1}{2} cos \frac{π}{8}

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option