જો $\tan x=\frac{3}{4}, \pi < x < \frac{3 \pi}{2},$ તો $\sin \frac{x}{2}, \cos \frac{x}{2}$ અને $\tan \frac{x}{2}$ નાં મૂલ્ય શોધો.
since $\pi < x < \frac{3 \pi}{2}, \cos x$ is negative.
Also $\quad \frac{\pi}{2}<\frac{x}{2}<\frac{3 \pi}{4}$
Therefore, $\sin \frac{x}{2}$ is positive and $\cos \frac{x}{2}$ is negative.
Now $\sec ^{2} x=1+\tan ^{2} x=1+\frac{9}{16}=\frac{25}{16}$
Therefore $\cos ^{2} x=\frac{16}{25}$ or $\cos x=-\frac{4}{5} \quad$ (Why?)
Now $2 \sin ^{2} \frac{x}{2}=1-\cos x=1+\frac{4}{5}=\frac{9}{5}$
Therefore $\sin ^{2} \frac{x}{2}=\frac{9}{10}$
or $\sin \frac{x}{2}=\frac{3}{\sqrt{10}} \quad$ (Why?)
Again $2 \cos ^{2} \frac{x}{2}=1+\cos x=1-\frac{4}{5}=\frac{1}{5}$
Therefore $\cos ^{2} \frac{x}{2}=\frac{1}{10}$or $\cos \frac{x}{2}=-\frac{1}{\sqrt{10}}(\text { Why? })$
Hence $\quad \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{3}{\sqrt{10}} \times\left(\frac{-\sqrt{10}}{1}\right)=-3$
રેડિયન માપ શોધો : $25^{\circ}$
જો $\sin x + {\sin ^2}x = 1,$ તો ${\cos ^8}x + 2{\cos ^6}x + {\cos ^4}x = $
રેડિયન માપ શોધો : $-47^{\circ} 30^{\prime}$
જો $x{\sin ^3}\alpha + y{\cos ^3}\alpha = \sin \alpha \cos \alpha $ અને $x\sin \alpha - y\cos \alpha = 0,$ તો ${x^2} + {y^2} = $
$2 \sin \left(12^{\circ}\right)-\sin \left(72^{\circ}\right)$ નું મુલ્ય ............ છે.