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Here are all the results I got

$$\tan(A+B+C)=a-b$$

And $$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$

And $$\cot A+\cot B + \cot C=0$$

How should I use these results?

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    $\begingroup$ Try finding the equation which has roots $t=1+\tan^2 A, u=1+\tan^2 B, v=1+\tan^2 C$ by setting $y=x^2+1$ and eliminating $x$ $\endgroup$ Apr 7 at 9:21
  • $\begingroup$ @MarkBennet You need the inverse function(s) as you want to input $1 + \tan^2 A$ into the function to end up with $\tan A$ as a root. I've added an answer based on your approach. $\endgroup$
    – Toby Mak
    Apr 7 at 9:52
  • $\begingroup$ @TobyMak I don't see why you need inverse functions at all. You are given the original cubic - the exercise can be done completely algebraically. $\endgroup$ Apr 7 at 10:00
  • $\begingroup$ @MarkBennet I thought you were referring to substituting $y = x^2 + 1$ into $x^3 - ax^2 + b = 0$. From your answer, it turns out you meant something else which clears my doubts up. $\endgroup$
    – Toby Mak
    Apr 7 at 10:04
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It is a trick. Let $p(x) = x^3-ax^2+b$, then $$p(x)= (x-\tan A)(x-\tan B)(x-\tan C)$$

Now you need $$p(i)\cdot p(-i)$$

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  • $\begingroup$ Is $i$ the complex number $i$ here? Can you tell me more about this, haven’t seen it before $\endgroup$
    – Aditya
    Apr 7 at 16:05
  • $\begingroup$ Yes it is $i^2=-1$. What do you need more? If $x_1,x_2,...x_n$ are roots for $p(x)$ then $$p(x) = a(x-x_1)(x-x_2)\cdots (x-x_n)$$ $\endgroup$
    – Aqua
    Apr 7 at 16:08
  • $\begingroup$ Ok I see that $p(i).p(-i)=1+(a+b)^2$, but how did you arrive at that? $\endgroup$
    – Aditya
    Apr 7 at 16:53
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Write $x=\tan A, y=\tan B, z=\tan C$.

By Vieta’s formula, we have $$x+y+z=a,\ \ xy+yz+zx=0,\ \ xyz=-b.$$

Now you want to find the value of $$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2.$$ Note that $$a^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx),$$ so $x^2+y^2+z^2=a^2$. Similarly, $$0=(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$ implies $x^2y^2+y^2z^2+z^2x^2=2ab$. Hence $$(1+x^2)(1+y^2)(1+z^2)=1+a^2+2ab+b^2=1+(a+b)^2.$$

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    $\begingroup$ It should be $xyz = - b$ but everything else is correct. $\endgroup$
    – Toby Mak
    Apr 7 at 9:28
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    $\begingroup$ @TobyMak Thanks for checking my answer! Fixed now :) $\endgroup$
    – Feng
    Apr 7 at 9:43
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So since others have shown workings I will do mine - Aqua's answer is equivalent and Toby Mak has a different way of doing the same thing.

Let $y=x^2+1$ so that $x^2=y-1$ and substitute as much as possible in the original cubic: $$x(y-1)-a(y-1)+b=0$$

Rearrange to isolate $x$ and obtain: $$x(y-1)=a(y-1)+b=ay-(a+b)$$

Now square this:

$$x^2(y-1)^2=(y-1)^3=\left(ay-(a+b)\right)^2$$ so that $$y^3-3y^2+3y+1-a^2y^2+2a(a+b)y-(a+b)^2=0$$

Now for the product of the roots you need just the constant term, and since the coefficient of $y^3$ is $+1$ you need the negative of the constant term, hence $(a+b)^2-1$

This uses only manipulation of the original equation and Vieta's relation in the final cubic.

This method, where applicable is completely algebraic, avoids working in any algebraic or analytic context beyond the world of the coefficients of the cubic, and avoids explicit manipulation of symmetric polynomials in the workings. I have always found these things an advantage. It uses no particular properties of the tan function, but I couldn't see how that would make a simpler solution. It only uses that the roots of two equations can be related.

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  • $\begingroup$ +1). Did you find this method yourself? Or did you see it in a book or somewhere? Thanks. $\endgroup$ Apr 7 at 10:25
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    $\begingroup$ @lonestudent It was in a book by WL Ferrar "Higher Algebra" or "Higher Algebra for Schools" which my maths master at school showed me (I think it may have been a combined edition). It was a rather old-fashioned text and had lots of exercises involving computation with things like roots of equations. I think that working with the roots of cubics may also have been on my A-level syllabus. There was little attempt really to get to grips with anything more abstract - though there was a hands-on proof that the elementary symmetric functions form a generating set for symmetric polynomials. $\endgroup$ Apr 7 at 10:41
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Set $x = 1 + y^2 \Rightarrow y = ±\sqrt{x - 1}$ so that $1 + \tan^2 A, 1 + \tan^2 B, 1 + \tan^2 C$ give the roots of $x^3 - ax^2 + b = 0$.

This gives $±(x-1)^{3/2} - a(x-1) +b=0$. By Vieta, the product of the roots is the constant term, but we do not have a polynomial yet. Squaring both sides of $±(x - 1)^{3/2} = a(x-1) - b$, the constant term is $(-1)^3 - (-a-b)^2 = -1 -(a+b)^2$ if we take the positive branch, and $(-a-b)^2 - (-1) = 1 + (a+b)^2$ if we take the negative branch. However, since $(1 + \tan^2 A)(1 + \tan^2 B)(1 + \tan^2 C)$ is always positive, the answer is $1 + (a+b)^2$.

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  • $\begingroup$ Notes: - The constant term can only be obtained from constant $\times$ constant, so we do not need to expand every single term. - For convenience, we move all the terms to the side where $x^3$ is positive, or else we have to divide by $a = -1$ to get the product for the negative branch. $\endgroup$
    – Toby Mak
    Apr 7 at 9:48
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It is basically an algebraic problem:

If $p,q,r$ are the roots of $$x^3-ax^2+b=0$$

it is sufficient to find the cubic equation whose roots are $w=p^2+1,y=q^2+1,z=r^2+1$

From $ap^2-b=p^3,$ we have $$(ap^2-b)^2=(p^3)^2=(p^2)^3$$

Replace $p^2$ with $w-1$ to find $$(a(w-1)-b)^2=(w-1)^3$$

$$\iff w^3+(\cdots)w^2+(\cdots)w-1-(a+b)^2=0$$

Obviously, we shall reach at the same equation if start with $y=q^2+1$ or $z=r^2+1$

Using Vieta's formula, $$wyz=-\dfrac{-1-(a+b)^2}1$$

See also :

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