sin ^2 frac{2 pi}{3}+cos ^2 frac{5 pi}{6}-tan | vedclass

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Question

(B) Given expression: $\sin ^2\left(\frac{2 \pi}{3}ight)+\cos ^2\left(\frac{5 \pi}{6}ight)-	an ^2\left(\frac{3 \pi}{4}ight)$ Using allied angle

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$1 / 2$

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(B) Given expression: $\sin ^2\left(\frac{2 \pi}{3}ight)+\cos ^2\left(\frac{5 \pi}{6}ight)-	an ^2\left(\frac{3 \pi}{4}ight)$ Using allied angles: $\sin(\pi-x) = \sin x$,$\cos(\pi-x) = -\cos x$,$	an(\pi-x) = -	an x$ $= \sin ^2\left(\pi-\frac{\pi}{3}ight)+\cos ^2\left(\pi-\frac{\pi}{6}ight)-	an ^2\left(\pi-\frac{\pi}{4}ight)$ $= \sin ^2\left(\frac{\pi}{3}ight)+\left(-\cos \frac{\pi}{6}ight)^2-\left(-	an \frac{\pi}{4}ight)^2$ $= \left(\frac{\sqrt{3}}{2}ight)^2+\left(-\frac{\sqrt{3}}{2}ight)^2-(-1)^2$ $= \frac{3}{4}+\frac{3}{4}-1$ $= \frac{6}{4}-1 = \frac{3}{2}-1 = \frac{1}{2}$

$\sin ^2 \frac{2 \pi}{3}+\cos ^2 \frac{5 \pi}{6}-\tan ^2 \frac{3 \pi}{4}=$

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