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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 .

Aakash Kumar
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3 Answers3

3

Hints:

  • $\dfrac\pi2<2<\pi$
  • the sum of the roots is $-3$
  • the product of the roots has sign opposite to $\tan 2$.
Bernard
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2

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 $\cot^{-1}\alpha + \cot^{-1}\frac1\alpha$.

  • I think in this question just knowing the sign would be better . – Aakash Kumar Jul 30 '16 at 16:27
  • @AakashKumar, I think you'll need to know more than just the sign. But the sign is needed, updating my answer now... –  Jul 30 '16 at 16:30
  • @ritwiksinha Answer is 1,2.4 – Aakash Kumar Jul 30 '16 at 16:31
  • @ritwiksinha I added missing info to my answer. –  Jul 30 '16 at 16:31
  • @tilper Did you mean $ \alpha $ by constant value . – Aakash Kumar Jul 30 '16 at 16:33
  • @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). –  Jul 30 '16 at 16:36
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    @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 $x<0$. – Zain Patel Jul 30 '16 at 16:36
  • @ZainPatel Indeed, thanks. I didn't expand on that method because this question only has a trig tag. –  Jul 30 '16 at 16:37
  • @tilper I dont doubt you but i got I calculated it for $-1$ and i got $-\pi/2$. –  Jul 30 '16 at 17:01
  • @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. –  Jul 30 '16 at 17:09
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$\alpha$ is a root of the equation then we have these two solutions $$x=-\frac{3}{2}-\frac{1}{2} \sqrt{9+4 \tan (2)}\lor x=\frac{1}{2} \sqrt{9+4 \tan (2)}-\frac{3}{2}$$ is this right? now we calculate $$\cot(\alpha)^{-1}+\cot(\frac{1}{\alpha})-\frac{\pi}{2}$$ is this right? then we get $$-\frac{\pi }{2}+\cot ^{-1}\left(\frac{1}{-\frac{3}{2}-\frac{1}{2} \sqrt{9+4 \tan (2)}}\right)-\cot ^{-1}\left(\frac{3}{2}+\frac{1}{2} \sqrt{9+4 \tan (2)}\right)$$ or $$-\frac{\pi }{2}-\cot ^{-1}\left(\frac{3}{2}-\frac{1}{2} \sqrt{9+4 \tan (2)}\right)+\cot ^{-1}\left(\frac{1}{\frac{1}{2} \sqrt{9+4 \tan (2)}-\frac{3}{2}}\right)$$ is this right? for both terms i got this value: $$-3.14159265358979323846264338327950288419716939937510582097494459230781640628620 8998628034825342117068$$

Dr. Sonnhard Graubner
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