Prove that:
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
$L.H.S$ $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$
$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$
$= R . H.S$
The incorrect statement is
Find the degree measures corresponding to the following radian measures ( Use $\pi=\frac{22}{7}$ ).
$\frac{7 \pi}{6}$
If $\sin \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1,$ then $\theta $ lies in which quadrant
$\sin \left( {\frac{\pi }{{10}}} \right)\sin \left( {\frac{{3\pi }}{{10}}} \right) = $
If $\sin A,\cos A$ and $\tan A$ are in $G.P.$, then ${\cos ^3}A + {\cos ^2}A$ is equal to