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

(N/A) We start with the $L.H.S$: $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
Substituting the standard trigonometric values: $\sin \frac{\pi}{6} = \frac{1}{2}$,$\cos \frac{\pi}{3} = \frac{1}{2}$,and $\tan \frac{\pi}{4} = 1$
$= (\frac{1}{2})^{2} + (\frac{1}{2})^{2} - (1)^{2}$
$= \frac{1}{4} + \frac{1}{4} - 1$
$= \frac{2}{4} - 1 = \frac{1}{2} - 1 = -\frac{1}{2}$
$= R.H.S$
Hence,the identity is proved.