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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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Prove : $\cot^2\large\frac{\pi}{6}$$+cosec \large\frac{5\pi}{6}$$+3\tan^2\large\frac{\pi}{6}$$=6$

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L.H.S
$\cot^2\large\frac{\pi}{6}$$+cosec \large\frac{5\pi}{6}$$+3\tan^2\large\frac{\pi}{6}$
$\Rightarrow \cot^2\large\frac{\pi}{6}$$+cosec(\pi- \large\frac{\pi}{6})$$+3\tan^2\large\frac{\pi}{6}$
$\Rightarrow (\cot\large\frac{\pi}{6})^2$$+cosec \large\frac{\pi}{6}$$+3(\tan\large\frac{\pi}{6})^2$
($cosec (\pi-\large\frac{\pi}{6})=$$cosec \large\frac{\pi}{6})$
$(\sqrt 3)^2+2+3\times (\large\frac{1}{\sqrt 3})^2$
$3+2+1=6$=R.H.S
Hence proved.
answered Apr 17, 2014 by sreemathi.v
 

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