Prove that :
$\cot ^{2} \frac{\pi}{6}+\cos ec \,\frac{5 \pi}{6}+3 \tan ^{2}\, \frac{\pi}{6}=6$
$L.H.S.$ $=\cot ^{2}\, \frac{\pi}{6}+\cos ec \,\frac{5 \pi}{6}+3 \tan ^{2}\, \frac{\pi}{6}$
$=(\sqrt{3})^{2}+\cos ec\, \left(\pi-\frac{\pi}{6}\right)+3\left(\frac{1}{\sqrt{3}}\right)^{2}$
$=3+\cos ec\, \frac{\pi}{6}+3 \times \frac{1}{3}$
$=3+2+1=6$
$= R . H.S$
$\cot x - \tan x = $
Prove that: $(\cos x+\cos y)^{2}+(\sin x-\sin y)^{2}=4 \cos ^{2} \frac{x+y}{2}$
If $\left| {\,a\,{{\sin }^2}\theta + b\sin \theta \cos \theta + c\,{{\cos }^2}\theta - \frac{1}{2}(a + c)\,} \right|\, \le \frac{1}{2}k,$ then ${k^2}$ is equal to
If $\sin \theta = - \frac{1}{{\sqrt 2 }}$ and $\tan \theta = 1,$ then $\theta $ lies in which quadrant
The incorrect statement is