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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions

Prove : $\sin^2\large\frac{\pi}{6}$$+\cos^2\large\frac{\pi}{3}$$-\tan^2\large\frac{\pi}{4}=-\frac{1}{2}$

1 Answer

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

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