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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $2\sin^2\large\frac{\pi}{6}+$$cosec ^2\large\frac{7\pi}{6}$$\cos^2\large\frac{\pi}{3}=\large\frac{3}{2}$

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L.H.S
$2\sin^2\large\frac{\pi}{6}+$$cosec ^2\large\frac{7\pi}{6}$$\cos^2\large\frac{\pi}{3}$
$\Rightarrow 2(\sin\large\frac{\pi}{6})^2+$$(cosec \large\frac{7\pi}{6})^2$$(\cos\large\frac{\pi}{3})^2$
$\Rightarrow 2(\large\frac{1}{2})^2+$$[cosec(\pi+\large\frac{\pi}{6})]^2($$\cos\large\frac{\pi}{3})^2$
$\Rightarrow 2\times \large\frac{1}{4}$$+(-cosec\large\frac{\pi}{6})^2\frac{1}{4}$
$cosec (\pi+\theta)=-cosec \theta$
$\Rightarrow \large\frac{1}{2}$$+2^2\large\frac{1}{4}=\frac{1}{2}$$+1=\large\frac{3}{2}$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 

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