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, $\tan \frac{x}{2}=2=\frac{2}{1}=\frac{\text{ perpendicular }}{\text{ base }}=\frac{AC}{AB}$
Using Pythagoras theorem, $(BC{)}^2-(AC{)}^2+(AB{)}^2$
$\Rightarrow (BC{)}^2=4+1=5$
$\Rightarrow BC=\sqrt{5}$
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.