$\alpha $ is a root of equation $x^2 + 3x \tan2 = 0$
Solution 1
Since you only asked for a hint, here's a big one.
$\cot^{1}\alpha + \cot^{1}\frac1\alpha$ is actually a constant. There are a few ways to prove this:
 Use the trig identity $\cot(x + y) = \frac{\cot x \cot y  1}{\cot x + \cot y}$ with $x = \cot^{1}\alpha$ and $y = \cot^{1}\frac1\alpha$.
 Use the inverse trig identity $\cot^{1}x + \cot^{1}y = \cot^{1}\left(\frac{xy  1}{x + y}\right)$ with $x = \alpha$ and $y = 1/\alpha$.
 Use calculus to show that the derivative of $\cot x + \cot\frac1x$ is zero.
The third method won't tell you what the constant value is but I include the method for thoroughness. I recommend using the second method. The first method is more or less the same as the second method, just with extra steps. Anyway, once you get the actual value of $\cot^{1}\alpha + \cot^{1}\frac1\alpha$, you'll immediately have the answer to your original question.
EDIT: Because of the rules of domains and ranges for inverse trig functions, the sign of $\alpha$ is important. Both roots of $x^2 + 3x  \tan 2$ are negative. Keep this in mind when evaluating $\cot^{1}\alpha + \cot^{1}\frac1\alpha$.
Solution 2
Hints:
 $\dfrac\pi2<2<\pi$
 the sum of the roots is $3$
 the product of the roots has sign opposite to $\tan 2$.
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Aakash Kumar
Updated on July 23, 2022Comments

Aakash Kumar 4 months
If $\alpha $ is a root of equation $x^2 + 3x \tan2 = 0$ then $\cot^{1} \alpha +\cot^{1} \frac{1}{\alpha} \frac{\pi}{2} $ cannot be equal to
 $ \frac{\pi}{2}$
 $ \frac{3\pi}{2}$
 $ \frac{\pi}{1}$
 $ 0$
For this I think I need to know the sign of $\alpha$ . Just a hint is sufficient .

Bernard over 6 years
cannot be equal to
what? 
Jean Marie over 6 years@Bernard cannot be equal to... one of the 4 answers given below.

lab bhattacharjee over 6 years

Dr. Sonnhard Graubner over 6 yearscomputing $\alpha$ and plugging this in the term below we get negativ results. Thus it cannot be answer 1,2,3,4

Aakash Kumar over 6 years@Dr.SonnhardGraubner Answer is 1,2,4

Dr. Sonnhard Graubner over 6 yearsis $\cot^{1}(x)$ equal to $arccot(x)$?

Aakash Kumar over 6 years@Dr.SonnhardGraubner yes

Admin over 6 yearsCotangent, not cosine.

Dr. Sonnhard Graubner over 6 yearsand why the $1$?

Admin over 6 yearsDunno, wasn't me.

Aakash Kumar over 6 yearsI think in this question just knowing the sign would be better .

Admin over 6 years@AakashKumar, I think you'll need to know more than just the sign. But the sign is needed, updating my answer now...

Aakash Kumar over 6 years@ritwiksinha Answer is 1,2.4

Admin over 6 years@ritwiksinha I added missing info to my answer.

Aakash Kumar over 6 years@tilper Did you mean $ \alpha $ by constant value .

Admin over 6 years@AakashKumar I don't understand the question. Did I mean $\alpha$ where? $\alpha$ has one of two possible values. The actual value of $\alpha$ doesn't matter. All that matters is that $\alpha < 0$ (and we know this because both possible values of $\alpha$ are negative). And $\cot^{1}\alpha + \cot^{1}\frac1\alpha$ is constant for $\alpha > 0$ and is a different constant for $\alpha < 0$ (as long as $\alpha$ and $1/\alpha$ are both in the domain of arccotangent).

Zain Patel over 6 years@tilper the third method will tell you that the function is constant, so using $x=1$ say will give you the constant for $x>0$ and $x=1$ will give you the constant for $x<0$.

Admin over 6 years@ZainPatel Indeed, thanks. I didn't expand on that method because this question only has a trig tag.

Admin over 6 years@tilper I dont doubt you but i got I calculated it for $1$ and i got $\pi/2$.

Admin over 6 years@ritwiksinha, $\cot^{1}(1) = 3\pi/4$. But note that there isn't really universal agreement for the range of arccotangent. Some people take it to be $(0,\pi)$ and some people take it to be $[\pi/2, 0) \cup (0, \pi/2]$. Only OP will know for sure which range OP needs to use, but I suspect it's $(0,\pi)$ because the other range means all 4 answer choices would be selected. And I guess that's technically a possibility but if I had to guess one way or the other, I wouldn't guess that all 4 are chosen.