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 $\quad \tan \frac{x}{2}=\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}=\frac{3}{\sqrt{10}} \times\left(\frac{-\sqrt{10}}{1}\right)=-3$
જો $A$ એ બીજા ચરણમાં હોય અને $3\tan A + 4 = 0,$ તો $2\cot A - 5\cos A + \sin A$ ની કિમત મેળવો.
જો $x + \frac{1}{x} = 2\cos \alpha $, તો ${x^n} + \frac{1}{{{x^n}}} = $
$\cot x - \tan x = $
જો $2y\,\cos \theta = x\sin \,\theta $ અને $2x\sec \theta - y\,{\rm{cosec}}\,\theta = 3,$ તો ${x^2} + 4{y^2} = $
$\sin \frac{x}{2}, \cos \frac{x}{2}$ અને $\tan \frac{x}{2}$ ની કિંમતો શોધો.: $\cos x=-\frac{1}{3}, x$ એ બીજા ચરણમાં છે.