Question

$\text{tan x}=-\frac{4}{3},\text{x\, in\, quadrant \,II.}$

Answer 1

Here, x is in quadrant II. $\frac{\pi}{2}<x<{\pi}$

i.e., $⇒\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$

Therefore, $sin\frac{x}{2},cos\frac{x}{2}\,and\,\,tan\frac{x}{2}\,are\,\,all\,positive.$

it is given that tan x=-$\frac{4}{3}.$

$sec^2x=1+tan^2x=1+(\frac{-4}{3})^2=1+\frac{16}{9}=\frac{25}{9}$

$cos^2x=\frac{9}{25}$

$⇒cos\,x=±\frac{3}{5}$

As x is in quadrant II, cos$$x is negative.

$∴cos\,x=\frac{-3}{5}$

Now, $cos x=2cos^2\,\frac{x}{2}=-1$

$⇒\frac{-3}{5}=2cos^2\frac{x}{2}-1$

$⇒2cos^2\frac{x}{2}-1-\frac{3}{2}$

$⇒2cos^2\,\frac{x}{2}=\frac{2}{5}$

$⇒2cos^2\frac{x}{2}=\frac{1}{5}$

$⇒cos\frac{x}{2}=\frac{1}{5}\,\,\,\\.[∴cos\,\frac{x}{2}\,is\,\,possitive]$

$=cos\frac{x}{2}=\frac{\sqrt5}{5}$

$sin^2+cos^2\,\frac{x}{2}=1$

$sin^2\frac{x}{2}+(\frac{1}{\sqrt5})^2=1$

$⇒sin^2\frac{x}{2}=1-\frac{1}{5}=\frac{4}{5}$

$⇒sin\frac{x}{2}=\frac{2}{\sqrt5}\,\,\,\,[∵\,sin\frac{x}{2}\,is\,positive]$

$∴ \,sin\frac{x}{2}=\frac{2\sqrt5}{5}$

$tan\frac{x}{2}=\frac{sin\frac{x}{2}}{cos\frac{x}{2}}=\frac{(\frac{2}{\sqrt5})}{(\frac{1}{\sqrt5})}=2$

Thus, the respective values of $sin\frac{x}{2},cos\frac{x}{2}\,and\,\,,tan\frac{x}{2}\,are\,\frac{2\sqrt5}{2},\frac{\sqrt5}{2},and\,2.$