If tan x = frac{3}{4} and pi

Question

(C) Given,$	an x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$.Since $\pi < x < \frac{3\pi}{2}$,dividing by $2$ gives $\frac{\pi}{2} < \frac{x}{2} <

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{10}}$

Vedclass · Answer

(C) Given,$	an x = \frac{3}{4}$ and $\pi Since $\pi In the second quadrant,$\cos 	heta$ is negative.We know that $\sec^2 x = 1 + 	an^2 x = 1 + (\frac{3}{4})^2 = 1 + \frac{9}{16} = \frac{25}{16}$.Therefore,$\sec x = \pm \frac{5}{4}$.Since $x$ is in the third quadrant,$\cos x$ must be negative,so $\cos x = -\frac{4}{5}$.Using the half-angle formula,$\cos^2 \frac{x}{2} = \frac{1 + \cos x}{2}$.Substituting the value of $\cos x$,we get $\cos^2 \frac{x}{2} = \frac{1 - \frac{4}{5}}{2} = \frac{1/5}{2} = \frac{1}{10}$.Since $\frac{x}{2}$ is in the second quadrant,$\cos \frac{x}{2}$ must be negative.Thus,$\cos \frac{x}{2} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

If $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3\pi}{2}$,then the value of $\cos \frac{x}{2}$ is

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