If L=\sin ^{2}(\frac{\\pi}{16})-\sin ^{2}(\frac{\\pi}{8}) and M=\cos ^{2}(\frac{\\pi }{16})-\sin ^{2}(\frac{\\pi}{8}), then : [5 Sep 2020 Shift 2]

Correct Answer is : M=12√2+12cosπ8

{M}={1}/{2 \sqrt{2}+\frac{1}{2} \cos \frac{\\pi}{8}

{L}={1}/{4 \sqrt{2}-\frac{1}{4} \cos \frac{\\pi}{8}

{M}={1}/{4 \sqrt{2}+\frac{1}{4} \cos \frac{\\pi}{8}

{L}=-{1}/{2 \sqrt{2}+\frac{1}{2} \cos \frac{\\pi}{8}

JEE Mains 05-Sep-2020 Shift 2 Solved Paper

Section: Mathematics

Question : 70 of 75

Marks: +1, -0

If L=sin2(

)−sin2(

) and
M=cos2(

)−sin2(

), then :

[5 Sep 2020 Shift 2]

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