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

Question

(D) Given $	an x = \frac{3}{4}$ and $\pi < x < \frac{3 \pi}{2}$.Since $x$ is in the third quadrant,$\sec^2 x = 1 + 	an^2 x = 1 + \frac{9}{16} = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given $	an x = \frac{3}{4}$ and $\pi Since $x$ is in the third quadrant,$\sec^2 x = 1 + 	an^2 x = 1 + \frac{9}{16} = \frac{25}{16}$.Thus,$\sec x = -\frac{5}{4}$ (as $\sec x Given $\pi In the interval $(\frac{\pi}{2}, \frac{3 \pi}{4})$,$\cos \frac{x}{2}$ is negative.Using the half-angle formula,$\cos \frac{x}{2} = -\sqrt{\frac{1 + \cos x}{2}}$.Substituting $\cos x = -\frac{4}{5}$,we get $\cos \frac{x}{2} = -\sqrt{\frac{1 - 4/5}{2}} = -\sqrt{\frac{1/5}{2}} = -\sqrt{\frac{1}{10}} = -\frac{1}{\sqrt{10}}$.

If $\tan x = \frac{3}{4}$ and $\pi < x < \frac{3 \pi}{2}$,then $\cos \frac{x}{2} = . . . . . .$

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