# Find    $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$    in  Q (8)   $\small \tan x = - \frac{4}{3}$        , x in quadrant  II

tan x = $-\frac{4}{3}$
We know that ,
$\sec^{2}x = 1 + \tan^{2}x$
$= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$  =  $\frac{25}{9}$
$sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in II quadrant  thats why sec x is -ve
So,

$sec x =-\frac{5}{3}$
Now,  $cos x = \frac{1}{\sec x}$  =  $-\frac{3}{5}$
We know that,
$cos x = 2\cos^{2}\frac{x}{2}- 1$                                                                               ($\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$  )
$-\frac{3}{5}+ 1 = 2$   $\cos^{2}\frac{x}{2}$

=  $\frac{-3+5}{5}$    =    $2\cos^{2}\frac{x}{2}$

$\frac{2}{5}$   =    $2\cos^{2}\frac{x}{2}$
$\cos^{2}\frac{x}{2}$ =  $\frac{1}{5}$
$\cos\frac{x}{2}$   = $\sqrt{\frac{1}{5}}$  = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of           $\cos\frac{x}{2}$     is +ve

$\cos\frac{x}{2}$   =  $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
we know that
$cos x =1 - 2\sin^{2}\frac{x}{2}$

$2\sin^{2}\frac{x}{2}$  =  1 -  $(-\frac{3}{5})$   =  $\frac{8}{5}$

$\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in II quadrant So value of sin x is +ve

$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$

$\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$

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