Find the values of other five trigonometric functions if $\cos x=-\frac{1}{2}, x$ lies in third quadrant.
$\cos x=-\frac{1}{2}$
$\therefore \sec x=\frac{1}{\cos x}=\frac{1}{\left(-\frac{1}{2}\right)}=-2$
$\sin ^{2} x+\cos ^{2} x=1$
$\Rightarrow \sin ^{2} x=1-\cos ^{2} x$
$\Rightarrow \sin ^{2} x=1-\left(-\frac{1}{2}\right)^{2}$
$\Rightarrow \sin ^{2} x=1-\frac{1}{4}=\frac{3}{4}$
$\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$
since $x$ lies in the $3^{\text {rd }}$ quadrant, the value of $\sin x$ will be negative.
$\therefore \sin x=-\frac{\sqrt{3}}{2}$
$\cos ec\, x=\frac{1}{\sin x}=\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}=-\frac{2}{\sqrt{3}}$
$\tan x=\frac{\sin x}{\cos x}=\frac{\left(-\frac{\sqrt{3}}{2}\right)}{\left(-\frac{1}{2}\right)}=\sqrt{3}$
$\cot x=\frac{1}{\tan x}=\frac{1}{\sqrt{3}}$
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.
If $\cos \theta = \frac{1}{2}\left( {x + \frac{1}{x}} \right)$, then $\frac{1}{2}\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = $
If $\sin (\alpha - \beta ) = \frac{1}{2}$ and $\cos (\alpha + \beta ) = \frac{1}{2},$ where $\alpha $ and $\beta $ are positive acute angles, then
If $\tan \theta = \frac{{20}}{{21}},$ cos$\theta$ will be