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Prove that:
$2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2}$
$2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{7 \pi}{6} \cos ^2 \frac{\pi}{3}=\frac{3}{2}$
Solution:
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Verified Answer
L.H.S. $=2 \sin ^2 \frac{\pi}{6}+\operatorname{cosec}^2 \frac{\pi}{3} \cos ^2 \frac{7 \pi}{6}$
$=2\left(\sin \frac{\pi}{6}\right)^2+\left(\operatorname{cosec} \frac{7 \pi}{6}\right)^2\left(\cos \frac{\pi}{3}\right)^2$
$=2\left(\frac{1}{2}\right)^2+\left[\operatorname{cosec}\left(\pi+\frac{\pi}{6}\right)\right]^2\left(\frac{1}{2}\right)^2$
$=2 \times \frac{1}{4}+\left(-\operatorname{cosec} \frac{\pi}{6}\right)^2\left(\frac{1}{4}\right)$
$[\because \operatorname{cosec}(\pi+\theta)=-\operatorname{cosec}]$
$=\frac{1}{2}+(2)^2 \frac{1}{4}=\frac{1}{2}+1=\frac{3}{2}=$ R.H.S.
$=2\left(\sin \frac{\pi}{6}\right)^2+\left(\operatorname{cosec} \frac{7 \pi}{6}\right)^2\left(\cos \frac{\pi}{3}\right)^2$
$=2\left(\frac{1}{2}\right)^2+\left[\operatorname{cosec}\left(\pi+\frac{\pi}{6}\right)\right]^2\left(\frac{1}{2}\right)^2$
$=2 \times \frac{1}{4}+\left(-\operatorname{cosec} \frac{\pi}{6}\right)^2\left(\frac{1}{4}\right)$
$[\because \operatorname{cosec}(\pi+\theta)=-\operatorname{cosec}]$
$=\frac{1}{2}+(2)^2 \frac{1}{4}=\frac{1}{2}+1=\frac{3}{2}=$ R.H.S.
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