If $\tan \theta = - \frac{1}{{\sqrt {10} }}$ and $\theta $ lies in the fourth quadrant, then $\cos \theta = $
$1/\sqrt {11} $
$ - 1/\sqrt {11} $
$\sqrt {\frac{{10}}{{11}}} $
$ - \sqrt {\frac{{10}}{{11}}} $
The equation ${\sin ^2}\theta = \frac{{{x^2} + {y^2}}}{{2xy}},x,y, \ne 0$ is possible if
Find the radius of the circle in which a central angle of $60^{\circ}$ intercepts an arc of length $37.4 \,cm$ ( use $\pi=\frac{22}{7}$ ).
Prove that:
$\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$
Find the radian measures corresponding to the following degree measures:
$240^{\circ}$
If the arcs of the same lengths in two circles subtend angles $65^{\circ}$ and $110^{\circ}$ at the centre, find the ratio of their radii.