Question

If $\tan x=-\frac{4}{3}$ and $x$ is in quadrant II, then which among the following is/are correct?
I. $\sin \frac{x}{2}=\frac{2 \sqrt{5}}{5}$
II. $\cos \frac{x}{2}=\frac{\sqrt{5}}{5}$
III. $\tan \frac{x}{2}=\frac{+2}{5}$

• A. I is correct
• B. II is correct
• C. I and II are correct
• D. All are correct

Answer 1

Answer: (C)

$\tan x=-\frac{4}{3}$
Given that, $x$ lies in second quadrant.
i.e., $\because \tan x=\frac{2 \tan \frac{x}{2}}{1-\tan ^2 \frac{x}{2}}=-\frac{4}{3} $
$\therefore 3 \tan \frac{x}{2}=-2\left(1-\tan ^2 \frac{x}{2}\right) $
$\Rightarrow 3 \tan \frac{x}{2}=-2+2 \tan ^2 \frac{x}{2}$
$\Rightarrow 2 \tan ^2 \frac{x}{2}-3 \tan \frac{x}{2}-2=0$
$ \Rightarrow 2 \tan ^2 \frac{x}{2}-(4-1) \tan \frac{x}{2}-2=0 $
$ \Rightarrow 2 \tan \frac{x}{2}\left(\tan \frac{x}{2}-2\right)+1\left(\tan \frac{x}{2}-2\right)=0 $
$ \Rightarrow \left(2 \tan \frac{x}{2}+1\right)\left(\tan \frac{x}{2}-2\right)=0 $
$\Rightarrow \tan \frac{x}{2}=2 \text { or } \tan \frac{x}{2}=-\frac{1}{2} $
$\therefore \frac{\pi}{2} < x < \pi \Rightarrow \frac{\pi}{4} < \frac{x}{2}<\frac{\pi}{2}$
i.e., $\frac{x}{2}$ lies in first quadrant.
Therefore, $\tan \frac{x}{2}=2=\frac{2}{1}=\frac{\text { perpendicular }}{\text { base }}=\frac{A C}{A B}$
Using Pythagoras theorem, $(B C)^2-(A C)^2+(A B)^2$
$\Rightarrow (B C)^2=4+1=5$
$ \Rightarrow B C=\sqrt{5}$
Now, $ \sin \frac{x}{2}=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{2}{\sqrt{5}}=\frac{2 \sqrt{5}}{5} $
$\therefore \cos \frac{x}{2}=\frac{\text { base }}{\text { hypotenuse }}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}$
Note Students are advised to use the positive and negative signs carefully according to quadrant system.