Russelle's triple tangent

We want to find the value of $\tan(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z).$ Using the well-known tangent addition formula:

For all real numbers $a, b,$ we have $\tan(a+b) = \dfrac{\tan a + \tan b}{1 - \tan a \tan b}$ (if these tangents are defined).

we have $\tan(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z)$ $= \dfrac{\tan(\tan^{-1}x+\tan^{-1}y) + z}{1 - \tan(\tan^{-1}x+\tan^{-1}y) \cdot z}$ $= \dfrac{\dfrac{x+y}{1-xy} + z}{1 - \dfrac{x+y}{1-xy} \cdot z}$ $= \dfrac{\dfrac{x+y}{1-xy} + z}{1 - \dfrac{x+y}{1-xy} \cdot z} \cdot \dfrac{1-xy}{1-xy}$ $= \dfrac{(x+y) + (1-xy)z}{(1-xy) - (x+y)z}$ $= \dfrac{x+y+z-xyz}{1-(xy+xz+yz)}$ By Vieta's Formulas, we have $x+y+z = \dfrac{799}{2},$ $xy+xz+yz = -\dfrac{400}{2} = -200,$ $xyz = \dfrac{1}{2},$ so $\dfrac{x+y+z-xyz}{1-(xy+xz+yz)} = \dfrac{\dfrac{799}{2} - \dfrac{1}{2}}{1 + 200} = \dfrac{399}{201} = \dfrac{133}{67}.$ The answer is $133+67 = \boxed{200}.$

We first establish the nature of the three roots. Let $f(u) = 2u^3-799u^2-400u-1$ . Then we find that $f(-1) = -401 < 0, f(-0.1) = 31.008 > 0, f(0) = -1 < 0$ . By the Intermediate Value Theorem, we know that there is a real root between $-1$ and $-0.1$ and another real root between $-0.1$ and $0$ . Also, we can apply Vieta's formula to find that $xyz = \frac{1}{2}, xy+yz+zx=\frac{-400}{2}=-200, x+y+z=\frac{799}{2}$ . Since the product of the three roots is a positive real number and we know that two of the roots are negative real numbers, the last root must be a positive real number. Without loss of generality, we let the roots be in the order $x < y < z$ .

We now use the inverse trigonometric identity $\tan^{-1} m + \tan^{-1} n = \tan^{-1} \frac{m+n}{1-mn}$ for $mn < 1$ . Since $x,y$ are both negative with magnitudes less than $1$ , then their product would be positive but less than $1$ . So we may apply the identity on $x, y$ and get $\omega = \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = \tan^{-1} \frac{x+y}{1-xy} + \tan^{-1} z$ . Since $x, y$ are both negative, their sum would be negative. So $x+y < 0$ and $xy < 1$ , and we can easily deduce that $\frac{x+y}{1-xy}$ is negative. But $z$ is positive, so we clearly have $\frac{x+y}{1-xy} \cdot z < 0 < 1$ , and we apply thus apply the identity on $\frac{x+y}{1-xy}, z$ to get $\omega = \tan^{-1} \frac{x+y}{1-xy} + \tan^{-1} z = \tan^{-1} \dfrac{\frac{x+y}{1-xy}+z}{1-\frac{x+y}{1-xy} \cdot z}$ . Manipulating this expression gives us $\omega = \tan^{-1} \frac{(x+y+z)-xyz}{1-(xy+yz+zx)}$ . We have already calculated earlier the values of $xyz, xy+yz+zx, x+y+z$ , so we can substitute these into the expression for $\omega$ we get $\omega = \tan^{-1} \frac{133}{67}$ . So we have $\tan \omega = \frac{133}{67}$ , and the answer we want is $133 + 67 = \boxed{200}$ .

By compound angle formulae, $tan(\omega)=tan(x+y+z)=\frac{x+y+z-xyz}{1-(xy+yz+zx)}$ .But by Vieta's formula, $x+y+z=399.5,xyz=0.5,xy+yz+xz=200$ . Substituting yields $tan(\omega) =\frac{133}{67}$ . Thus, $a+b=200$ .

We must first prove that all the 3 roots real.

Let $f(u) = 2u^3 - 799u^2 - 400u - 1 = 0$ .

$f( - \frac {1}{2} < 0, f( - \frac {1}{3} > 0, f(0) < 0, f(401) > 0$ .

So there exist at least one root between $(- \frac {1}{2}, - \frac {1}{3}), (- \frac {1}{3}, 0), (0, 401)$ . Hence, all roots are real.

Thus, $\arctan (\tan x) = x, \arctan (\tan y) = y, \arctan (\tan z) = z$ are real and valid.

By Vieta's formula, $x+y+z = \frac {799}{2}, xy+xz + yz=-200, xyz = \frac {1}{2}$

Because $\omega = \arctan x + \arctan y + \arctan z$

$\tan \omega = \tan ( \arctan x + \arctan y + \arctan z )$

By Compound angle formula

$\tan \omega = \tan ( (\arctan x + \arctan y ) + \arctan z )$

$\LARGE \tan \omega = \frac { \frac {x + y}{1 -xy} + z } { 1 - \frac {x + y}{1 -xy} \cdot z }$

$\large \tan \omega = \frac { (x + y + z) - (xyz) } {1 - (xy+xz+yz) }$

Substitute the values gives

$\tan \omega = \frac {133}{67} \Rightarrow a=133,b=67 \Rightarrow a+b=\boxed{200}$

Note the identity $$\tan (a + b + c) = \frac{\tan a + \tan b + \tan c - \tan a \tan b \tan c}{1 - (\tan a \tan b + \tan b \tan c + \tan c \tan a)}$$ Take $a = \arctan x$ , $b = \arctan y$ and $c = \arctan z$ , then $$\tan \Big(\arctan x + \arctan y + \arctan z\Big) = \tan \omega = \frac{x + y + z - xyz}{1 - (xy + yz + zx)}$$ By Viete's formulae we have $$x + y + z = \frac{799}{2}$$ $$xy + yz + zx = \frac{-400}{2} = -200$$ $$xyz = \frac{1}{2}$$ Hence $$\tan \omega = \frac{\frac{799}{2} - \frac{1}{2}}{1 + 200} = \frac{133}{67}$$ So $$a + b = 133 + 67 = \boxed{200}$$

[Without loss of generality, the values of $x, y,$ and $z$ lie within $\mathbb{R}$ . By the properties of tangent and arctangent, therefore, $tan(tan^{-1}(x)) = x$ .]

Let us focus on the equation involving $\omega$ first. For a simpler equation to work with, let $\alpha = tan^{-1}(x) + tan^{-1}(y)$ and $\beta = tan^{-1}(z)$ . Hence, $\omega = \alpha + \beta$ .

By the formula for the tangent of a sum, we have $tan(\alpha) = \frac {x + y} {1 - xy}$ and $tan(\omega) = \frac {tan(\alpha) + tan(\beta)} {1 - tan(\alpha)tan(\beta)}$ . By substitution and simplification, we have:

$tan(\omega) = \frac {(x + y + z) - (xyz)} {1 - (xy + yz + zx)}$ .

And we go back to the polynomial expression, where we can use Vieta's formula to find its roots. For the roots $x, y,$ and $z$ of this polynomial, the following equations hold:

$(1) x + y + z = \frac {799} {2} \\ (2) xy + yz + zx = \frac {-400} {2} \\ (3) xyz = \frac {1} {2}$

Substituting these values into the expression for $tan(\omega)$ , and simplifying, we have $tan(\omega) = \frac {399} {201} = \frac {133} {67}$ , and the required answer is $\boxed {a + b = 133 + 67 = 200}$ .

Here’s an approach using complex numbers and polynomial factorization, instead of trigs.

Define cubic polynomial $f(u) := 2u^3 - 799u^2 - 400u - 1 = 2(u - x)(u - y)(u - z).$

Consider the complex numbers $1 + ix, 1+iy, 1+iz$ whose arguments are $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ , respectively.

Then $\omega$ is the argument of the product of these complex numbers (up to a multiple of $2\pi$ ): $\begin{aligned} \underbrace{(1+ix)(1+iy)(1+iz)}_{\text{factor out } -i \text{ for each term}} &= (-i)^{3}(i-x)(i-y)(i-z) \\ &= \frac{i}{2} \cdot 2(i-x)(i-y)(i-z) \\ &= \frac{i}{2} \cdot f(i) \\ &= \frac{i}{2}(798 - 402i) \\ &= 201 + 399i, \end{aligned}$ namely, $\tan \omega = \frac{399}{201} = \frac{133}{67}.$ Thus, $a+b = 133 + 67 = \boxed{\mathbf{200}}.$

First we will show that the roots are indeed all real. To do so we take $f(u) =\:$ $2u^{3} - 799u^{2} - 400u -1$ . So $f'(u) =\:$ $6u^{2} - 1598u - 400$ and it's discriminant is obviously greater than zero $\implies\:$ $f'(u)$ has all real roots $\implies\:$ $f(u)$ has all it's roots as real $\big($ Using Intermediate-Value Theorem and the continuity of f(u) $\big)$ .

Now using Vieta's Formula we have, $x + y + z =\:$ $\displaystyle\frac{799}{2}$ $, xy + yz + zx =\:-200$ $,xyz =\:$ $\displaystyle\frac{1}{2}$ .

We know from inverse trigonometry $\tan^{-1}x + tan^{-1}y =\:$ $\displaystyle\tan^{-1}\frac{x + y}{1-xy}$ , extending it we obtain $\omega =\:$ $\displaystyle\tan^{-1}\frac{x + y + z - xyz}{1 - xy - yz - zx}$ $=\:$ $\displaystyle\tan^{-1}\frac{\frac{799}{2} - \frac{1}{2}}{1 + 200}$

$\implies\:$ $\omega =\:$ $\displaystyle\tan^{-1}\frac{133}{67}$

$\implies\:$ $\tan\omega=\:$ $\frac{133}{67}$ $=\:$ $\frac{a}{b}$

So our answer is $a + b =\:$ $133 + 67 =\:$ $\boxed{200}$

If we wish to find $\tan\omega$ , we need to first find a formula for $\tan(a + b + c)$ . Through some basic knowledge of trig identities and algebraic manipulation, we obtain $\tan(a + b + c) = \dfrac{\tan{a} + \tan{b} + \tan{c} - \tan{a}\tan{b}\tan{c}}{1 - \tan{a}\tan{b} - \tan{a}\tan{c} - \tan{b}\tan{c}}$ Now we have $tan\omega = tan(\tan^{-1}x + \tan^{-1}y + \tan^{-1}z) = \dfrac{x + y + z - xyz}{1 - xy - yz - xz}$ Now we have to find the sum of the roots, the product of the roots, and the sum of the pairwise products of the roots. This can be done simply using Vieta's formulas: $x + y + z = \dfrac{-a_{n - 1}}{a_n}$ $xyz = (-1)^n \dfrac{a_0}{a_n}$ $xy + yz + xz = \dfrac{a_{n - 2}}{a_n}$ Now we have $tan\omega = \dfrac{\dfrac{799}{2} - \dfrac{1}{2}}{1 + \dfrac{400}{2}}$ which simplifies to $\dfrac{133}{67}$ , which is our final answer. a + b + 133 + 67 = 200.

We know that, $x$ , $y$ and $z$ are the real roots of the cubic equation, so,
$x$ + $y$ + $z$ = - $\frac{-799}{2}$ .....................(1)
$x$ $y$ + $y$ $z$ + $z$ $x$ = $\frac{-400}{2}$ ........................(2)
$x$ $y$ $z$ = - $\frac{-1}{2}$ ........................(3)
Now, $tan$ ω = $tan$ ( $tan^{-1}$ $x$ + $tan^{-1}$ $y$ + $tan^{-1}$ $z$ )

⇒ $tan$ ω = $\frac{tan(tan^{-1}x) + tan(tan^{-1}y) + tan(tan^{-1}z) - tan(tan^{-1}x).tan(tan^{-1}y).tan(tan^{-1}z)}{1 - tan(tan^{-1}x).tan(tan^{-1}y) - tan(tan^{-1}y).tan(tan^{-1}z) - tan(tan^{-1}z).tan(tan^{-1}x)}$

⇒ $tan$ ω = $\frac{x + y + z - xyz}{1 - xy - yz - zx}$

⇒ $tan$ ω = $\frac{-\frac{-799}{2} - (-\frac{-1}{2})}{1 - \frac{-400}{2}}$ [ By putting the value from the equations (1), (2) and (3)]

⇒ $tan$ ω = $\frac{133}{67}$
So, the value of $a$ + $b$ = $133$ + $67$ = $200$ [ANSWER]

We use the arctan addition formula to simplify, as follows.

$\implies \arctan (x) + \arctan (y) = \arctan \left( \dfrac {x+y}{1-xy} \right)$

$\implies \arctan (x) + \arctan (y) + \arctan (z) = \arctan \left( \dfrac {x+y}{1 - xy} \right) + \arctan \left( z \right)$

Again, we apply the arctan formula and we simplify. We get that $\arctan (x) + \arctan (y) + \arctan (z) = \arctan \left( \dfrac {xyz - x - y - z}{xy + xz + yz - 1} \right).$ Therefore, $\tan \omega = \dfrac {xyz - \left( x+y+z \right)}{xy+xz+yz-1}.$ We use Vieta's formulae on $2u^3 - 799u^2 - 400u - 1 = 0.$ $\implies xyz = \dfrac {1}{2}$ , since it is the product of the roots.

$\implies x + y + z = \dfrac {799}{2}$ , since it is the sum of the roots.

$\implies xy + xz + yz = -200$ , since it is the cyclic 2-product of the roots.

Therefore, $\tan \omega = \dfrac {\dfrac {1}{2} - \dfrac {799}{2}}{-201} = \dfrac {133}{67},$ so our answer is $133+67=\boxed{200}$ .

By VIETA'S FORMULA, $x+y+z$ = $799/2$ ..... $xy+yz+xz$ = $-400/2$ = $-200$ ....... $xyz$ = $1/2$ ...... $w$ = $A+B+C$ where $A$ = $arctan$ x , $B$ = $arctan$ y , $C$ = $arctan$ z ....... From above we can conclude that $tanA$ = x , $tanB$ = y , $tanC$ = z ............. Now, $tanw$ = $tan (A+B+C$ ) i.e.[ $tan A$ + $tan B$ + $tan C$ - $tanA$ $tanB$ $tanC$ ] $/$ [ $1- (tanA$ $tanB$ + $tanB$ $tanC$ + $tanA$ $tanC$ ) ] .............. which is [ $x+y+z$ - $xyz$ ] $/$ [ $1 - xy-yz-xz$ ]. { We had found the value of these expressions using vieta's formula. } =[ $799/2$ - $1/2$ ] $/$ [ $1-(-200)$ ] i.e. =[ $399$ ] $/$ [ $201$ ] = [ $133$ ] $/$ [ $67$ ] which is in the form $^a/_b$ .............. Now, $a+b$ = $133+67$ i.e. $200$ .

Let $\tan^{-1}x=a ,\tan^{-1}y=b ,\tan^{-1}z=c$ so that $\tan a=x ,\tan b=y ,\tan c=z$ . Then, we have $\omega=a+b+c$ which equivalent with $\tan\omega=\tan(a+b+c)$ . By using trigonometric identities, we get $\tan(a+b+c)=\frac{\tan a+\tan b+\tan c-\tan a.\tan b.\tan c}{1-(\tan a.\tan b+\tan b.\tan c+\tan c.\tan a)}$ which equivalent with $\tan\omega=\frac{x+y+z-xyz}{1-(xy+yz+zx)}$ Based on Vieta's Formula, we can conclude that $x+y+z=\frac{799}{2},xy+yz+zx=-200,xyz=\frac{1}{2}$ Hence, we get $\tan\omega=\frac{\frac{799}{2}-\frac{1}{2}}{1+200}=\frac{399}{201}=\frac{133}{67}$ We know that $133$ and $67$ are coprime which giving the answer $a+b=200$ .

We try to simplify the expression $\tan(\tan^{-1}x+\tan^{-1}y+\tan^{-1}z)$ . Using tangent addition: $\tan(\tan^{-1}x+\tan^{-1}y)=\dfrac{x+y}{1-xy}$ We can then say: $\tan(\tan^{-1}\dfrac{x+y}{1-xy}+\tan^{-1}z)=\dfrac{\dfrac{x+y}{1-xy}+z}{1-\dfrac{z(x+y)}{1-xy}}=\dfrac{x+y+z-xyz}{1-xy-xz-yz}$ Using Vieta, $x+y+z=\dfrac{799}{2}$ $xy+xz+yz=-200$ $xyz=\dfrac{1}{2}$ Now, our expression is equal to $\dfrac{\dfrac{799}{2}-\dfrac{1}{2}}{1+200}=\dfrac{399}{201}=\dfrac{133}{67}$ The answer is $133+67=\boxed{200}$ .

Suppose $p=\tan^{-1}x , q=\tan^{-1}y , r=\tan^{-1}z$ .

Therefore, $\tan \omega=\tan(p+q+r)$

= $\frac{\tan p +\tan q+\tan r - \tan p \tan q\tan r}{1-(\tan p \tan q+\tan r\tan p +\tan q\tan r) }$

= $\frac{x+y+z-xyz}{1-(xy+yz+zx)}$ .

According to Vieta's Formulas: $x+y+z=\frac{799}{2}, xy+yz+zx=-200, xyz=\frac{1}{2}$ , so $\tan \omega=\frac{\frac{799}{2}-\frac{1}{2}}{1-(-200)}=\frac{133}{67}$ .

The answer is: 133+67=200.

We have, tanw

= tan( $tan^{-1}x$ + $tan^{-1}y$ + $tan^{-1}z$ )

=tan( tan^{-1} \(\frac{x+y+z-xyz}{1-xy-yz-zx} )

We know,

x + y + z = $\frac{799}{2}$

xyz = 1

xy + yz + zx = -200

Putting successively we get,

= $\frac{399}{201}$

= $\frac{133}{67}$

So a+b = 67+133 = 200 (ans)

By information in question, x + y + z = 799/2 , xy + yz + zx = -200 , xyz = 1/2 .............. (i)

w = tan^{-1}(x) + tan^{-1}(y) + tan^{-1}(z)

= tan^{-1}[(x+y)/(1-xy)] + tan^{-1}(z) ........... (Using inverse trigonometric identities)

= tan^{-1}[(x+y+z - xyz) / (1-(xy+yz+zx))]

Now, put values from (i) which gives

tan(w) = 399/201 = 133/67.

Clearly, a = 133, b = 67 and a + b = 200

as we know arctan x+arctan y+arc tan z= arctan[x+y+z-xyz/(1-xy-yz-zx)]=w(given) =>tan w=[x+y+z-xyz/(1-xy-yz-zx)]...........(1) now given equation is 2u^3−799u^2−400u−1=0 as x y and z are= roots (as comparing coefficients with ax^3+bx^2+cx+d) =>sum of roots(-b/a) or x+y+z=799/2.........(2) sum of product of roots(c/a) or xy+yz+zx=-200...............(3) now product of roots(-d/a) or xyz=1/2...........(4) so put all values in (1) we get tan w=399/201 (as a and b are co prime) then tan w=133/67 a+b=133+67=200

We have that $\tan(a+b+c) = \frac{\tan(a)+\tan(b)+\tan(c) - \tan(a) \tan(b)\tan(c)}{1-(\tan(a)\tan(b)+\tan(a)\tan(c)+\tan(b)\tan(c))}$ , and by Vieta's Formulas, $x+y+z = \frac{799}{2}\\ xyz = \frac{1}{2} \\ xy+xz+yz = -200$

So, we have $\tan(\omega) = \frac{\frac{799}{2}-\frac{1}{2}}{1-(-200)} = \frac{133}{67}$ .

〖w= tan〗^(-1)⁡〖x +tan^(-1)⁡x+ tan^(-1)⁡z 〗

w =〖 tan〗^(-1)⁡〖(x+y)/(1-xy)〗+ tan^(-1)⁡z

w =〖 tan〗^(-1)⁡〖(x+y+z-xyz)/(1-(xy+yz+zx))〗

We know that x+y+z= 799/2

xy+yz+zx= (-400)/2= -200

xyz= 1/2

Putting these values, we get w = tan^(-1)⁡〖133/67〗

tan⁡〖w= 133/67〗= a/b

Therefore, a + b = 133 + 67 = 200

By $Vieta's formula$ , we know that

$x + y + z = \frac {799}{2}$ ;

$x \cdot y + y \cdot z + z \cdot x = -200$ ;

$x \cdot y \cdot z = \frac {1}{2}$ ;

Now, suppose $\tan^{-1} x = p$ , $\tan^{-1} y = q$ and $\tan^{-1} z = r$ .

$\Rightarrow \tan p = x$ , $\tan q = y$ and $\tan r = z$ .

Therefore, $\omega = \tan^{-1} x + \tan^{-1} y + \tan^{-1} z = p + q + r$ ,

and

$\tan \omega = \tan (p + q + r)$ .

In such cases where we have to use $expanded$ form of $\tan (\theta_1 + \theta_2 + \theta_3 \ldots + \theta_n)$ , it is good to remember a formula derived:

$\boxed {\tan (\theta_1 + \theta_2 + \theta_3 \ldots + \theta_n) = \frac {S_1 - S_3 + S_5 \ldots \pm S_n}{1 - S_2 + S_4 \ldots \mp S_{n-1}}}$

where

$S_1$ = sum of tan of angles taken $one$ at a time = $\tan \theta_1 + \tan \theta_2 + \tan \theta_3 + \ldots + \tan \theta_n$

$S_2$ = sum of tan of angles taken $two$ at a time = $\tan \theta_1 \cdot \tan \theta_2 + \tan \theta_2 \cdot \tan \theta_3 + \dots$

and similarly, $S_3 , S_ 4 \ldots S_n$ .

$\Rightarrow, \tan \omega = \frac { (\tan p + \tan q + \tan r) - (\tan p \cdot \tan q \cdot \tan r)}{1 - (\tan p \cdot \tan q + \tan q \cdot \tan r + \tan r \cdot \tan p)}$ .

Therefore, $\tan \omega = \frac {(x + y + z) - (x \cdot y \cdot z)}{1 - (x \cdot y + y \cdot z + z \cdot x)} = \frac {\frac {799}{2} - \frac {1}{2}}{1 - (-200)} = \frac {399}{201} = \frac {133}{67}$

Hence, $\tan \omega = \frac {133}{67} = \frac {a}{b}$

$\Rightarrow \boxed {a = 133} , \boxed {b = 67}$ and $a + b = \boxed {200}$

Let $\alpha = \tan^{-1} x, \beta = \tan^{-1} y, \gamma = \tan^{-1} z$ . Then $\omega = \alpha + \beta + \gamma$ . $\begin{aligned} \tan \omega &= \tan(\alpha + \beta + \gamma) \\ &= \frac{\tan \alpha + \tan(\beta + \gamma)}{1 - \tan\alpha \tan(\beta + \gamma)} \\ &= \frac{\tan\alpha + \frac{\tan\beta + \tan\gamma}{1 - \tan\beta \tan\gamma}}{1 - \tan\alpha \frac{\tan\beta + \tan\gamma}{1 - \tan\beta \tan\gamma}} \\ &= \frac{\tan\alpha (1 - \tan\beta \tan\gamma) + \tan\beta + \tan\gamma}{1 - \tan\beta \tan\gamma - \tan\alpha(\tan \beta + \tan\gamma)} \\ &= \frac{x + y + z - xyz}{1 - yz - xy - xz} \end{aligned}$ Since $x,y,z$ are roots of $2u^3 - 799u^2 - 400u - 1 = 0$ , $\begin{aligned} \frac{799}{2} &= x + y + z\\ -200 &= xy + xz + yz \\ \frac{1}{2} &= xyz \end{aligned}$ Hence $\tan\omega = \frac{\frac{799}{2} - 1}{1 - (-200)} = \frac{133}{67}$

Let A = arctan x , B = arctan y, and C = arctan z If arctan x = A, then x = tan A If arctan y = B, then y = tan B If arctan z = C, then z = tan C Therefore, tan w = tan (A + B + C) Further simplification using tangent formula gives, tan w = ((tan A + tan B + tan C) - (tan A)(tan B)(tan C))/(1 - (tan A + tan B) - (tan A + tan C) - (tan B + tan C)) which is equal to (x + y + z - xyz)/(1 - xy - xz - yz) or by Vieta's Theorem (799/2 - 1/2)/(1 - -400/2) = (798/2)/(402/2) = 399/201 = 133/67 Therefore a = 133 and b = 67 a + b = 133 + 67 = 200

We know that $\arctan(x)+\arctan(y)+\arctan(z)=\arctan(\frac{x+y}{1-xy})+\arctan(z)$ This is equal to $\arctan(\frac{z+\frac{x+y}{1-xy}}{1-\frac{z(x+y)}{1-xy}}$

Simplifying, we get: $(1-xy)*\frac{z+\frac{x+y}{1-xy}}{1-xy-z(x+y} \Longrightarrow \frac{x+y+z-xyz}{1-xy-xz-yz}$ By Vieta's, we can substitute in the values: $\frac{799/2-1/2}{1-(-200} \Longrightarrow {399/201} \Longrightarrow \frac{133}{67}$
Therefore, the answer is $\boxed{200}$

For a polynomial ax^3 +bx^2 +cx+d=0 Adding the roots gives -b/a Here 799/2

Multiplying the roots gives -d/a Here,1/2

We also get xy+yz+zx = c/a Here it is,(-400/2)

So now using the inverse circular formula :-

omega=tan^(-1)[[x+y+z-(xyz)]/[1-(xy+yz+zx)]] Solving we get:- omega=tan^(-1)[133/67] Now, tan(omega)=133/67. Hence a+b=200

Using the addition formula, $\tan (a+b+c) = \frac{ \tan a + \tan b + \tan c -\tan a \tan b \tan c }{1-( \tan a \tan b + \tan b \tan c + \tan c \tan a )}$ . Since $ω= \tan ^{-1} x+ \tan ^{-1} y+ \tan ^{-1} z$ , let $a= \tan ^{-1} x, b= \tan ^{-1} y, c= \tan ^{-1} z$ , $x= \tan a , y= \tan b , z= \tan c$ , therefore $w=a+b+c$ , $\tan w$ = $\tan (a+b+c)$ = $\frac{ \tan a + \tan b + \tan c -\tan a \tan b \tan c }{1-( \tan a \tan b + \tan b \tan c + \tan c \tan a )}$ = $\frac{x+y+z-xyz}{1-(xy+yz+zx)}$ . For a cubic equation $ax^3+bx^2+cx+d=0$ , the roots are denoted as $\alpha , \beta , \gamma$ , then $\alpha + \beta + \gamma = -\frac{b}{a}$ , $\alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a}$ , and $\alpha \beta \gamma = -\frac{d}{a}$ . Since x, y and z are the real roots of $2u^3-799u^2-400u-1=0$ , $x+y+z=-\frac{-799}{2}=\frac{799}{2}$ , $xy+yz+zx=\frac{-400}{2}=-200$ , $xyz=-\frac{-1}{2}=\frac{1}{2}$ . Therefore, $\tan w = \frac{ \frac{799}{2} - \frac{1}{2}}{1-(-200)} = \frac{399}{201} = \frac{133}{67} , 133+67=200$ .