Prove that:

$2 \sin ^{2} \frac{\pi}{6}+\cos ec ^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}$

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$L.H.S.$ $=2 \sin ^{2} \frac{\pi}{6}+\cos ec ^{2}\, \frac{7 \pi}{6} \,\cos ^{2} \,\frac{\pi}{3}$

$=2\left(\frac{1}{2}\right)^{2}+\cos ec ^{2}\left(\pi+\frac{\pi}{6}\right)\left(\frac{1}{2}\right)^{2}$

$=2 \times \frac{1}{4}+\left(-\cos ec \,\frac{\pi}{6}\right)^{2}\left(\frac{1}{4}\right)$

$=\frac{1}{2}+(-2)^{2}\left(\frac{1}{4}\right)$

$=\frac{1}{2}+\frac{4}{4}=\frac{1}{2}+1=\frac{3}{2}$

$=R .H.S.$

Similar Questions

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$\frac{7 \pi}{6}$

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  • [IIT 2020]