Question.
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
solution:
L.H.S. $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$
$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$
$=$ R.H.S.
L.H.S. $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$
$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$
$=$ R.H.S.
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