# $\alpha$ is a root of equation $x^2 + 3x -\tan2 = 0$

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## 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, 2022

• Aakash Kumar over 1 year

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

1. $\frac{\pi}{2}$
2. $\frac{3\pi}{2}$
3. $\frac{\pi}{1}$
4. $0$

For this I think I need to know the sign of $\alpha$ . Just a hint is sufficient .

• Bernard over 7 years
cannot be equal to what?
• Jean Marie over 7 years
@Bernard cannot be equal to... one of the 4 answers given below.
• lab bhattacharjee over 7 years
• Dr. Sonnhard Graubner over 7 years
computing $\alpha$ and plugging this in the term below we get negativ results. Thus it cannot be answer 1,2,3,4
• Aakash Kumar over 7 years
• Dr. Sonnhard Graubner over 7 years
is $\cot^{-1}(x)$ equal to $arccot(x)$?
• Aakash Kumar over 7 years
@Dr.SonnhardGraubner yes
Cotangent, not cosine.
• Dr. Sonnhard Graubner over 7 years
and why the $-1$?
Dunno, wasn't me.
• Aakash Kumar over 7 years
I think in this question just knowing the sign would be better .
@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 7 years
@tilper Did you mean $\alpha$ by constant value .
@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).
@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 $x<0$.
@tilper I dont doubt you but i got I calculated it for $-1$ and i got $-\pi/2$.
@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.