Prove that:
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
$L.H.S$ $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$
$=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}$
$=\frac{1}{4}+\frac{1}{4}-1=-\frac{1}{2}$
$= R . H.S$
$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ is equal to
Find the value of $\cos \left(-1710^{\circ}\right)$.
Which of the following is correct
Find $\sin \frac{x}{2}, \cos \frac{x}{2}$ and $\tan \frac{x}{2},$ if $\tan x=\frac{-4}{3}, x$ in quadrant $II$
If $\tan \,(A - B) = 1,\,\,\,\sec \,(A + B) = \frac{2}{{\sqrt 3 }},$ then the smallest positive value of $B$ is