Celebration problem marking the end of my exams

From

$\begin{aligned}\tan\left(3\tan^{-1}\dfrac14\right)=&\dfrac{3\tan\left(\tan^{-1}\dfrac14\right)-\tan^3\left(\tan^{-1}\dfrac14\right)}{1-3\tan^2\left(\tan^{-1}\dfrac14\right)}=\dfrac{3\left(\dfrac14\right)-\left(\dfrac14\right)^3}{1-3\left(\dfrac14\right)^2}\\=&\dfrac{\dfrac34-\dfrac1{64}}{1-\dfrac3{16}}=\dfrac{48-1}{64}\times\dfrac{16}{13}=\dfrac{47}{52}\end{aligned}$

From Equation

$\begin{aligned}3\tan^{-1}\left(\dfrac14\right)+\tan^{-1}\left(\dfrac1{20}\right)=&\dfrac{\pi}4-\tan^{-1}\left(\dfrac1x\right)\\\tan\left(3\tan^{-1}\dfrac14+\tan^{-1}\dfrac1{20}\right)=&\tan\left(\dfrac{\pi}4-\tan^{-1}\dfrac1x\right)\\\dfrac{\tan\left(3\tan^{-1}\dfrac14\right)+\tan\left(\tan^{-1}\dfrac1{20}\right)}{1-\tan\left(3\tan^{-1}\dfrac14\right)\tan\left(\tan^{-1}\dfrac1{20}\right)}=&\dfrac{\tan\left(\dfrac{\pi}4\right)-\tan\left(\tan^{-1}\dfrac1x\right)}{1+\tan\left(\dfrac{\pi}4\right)\tan\left(\tan^{-1}\dfrac1x\right)}\\\dfrac{\dfrac{47}{52}+\dfrac1{20}}{1-\left(\dfrac{47}{52}\right)\left(\dfrac1{20}\right)}=&\dfrac{1-\dfrac1x}{1+\dfrac1x}\\\dfrac{235+13}{260}\times\dfrac{1040}{993}=&\dfrac{x-1}x\times\dfrac x{x+1}\\\dfrac{1984}{1986}=&\dfrac{x-1}{x+1}\\\dfrac{1985-1}{1985+1}=&\dfrac{x-1}{x+1}\\x=&\boxed{1985}\end{aligned}$

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Notes:

$\tan(\tan^{-1}x)=x$

$\tan(3x)=\dfrac{3\tan{x}-\tan^3x}{1-3\tan^2x}$

$\tan(x+y)=\dfrac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}$

$\tan(x-y)=\dfrac{\tan{x}-\tan{y}}{1+\tan{x}\tan{y}}$

$\begin{aligned} \frac \pi 4 - \tan^{-1} \frac 1x & = 3\tan^{-1} \frac 14 + \tan^{-1} \frac 1{20} \\ \tan^{-1} \frac {1-\frac 1x}{1+\frac 1x} & = \tan^{-1} \frac {\frac 14\left(3-\left(\frac 14 \right)^2\right)}{1-3\left(\frac 14\right)^2} + \tan^{-1} \frac 1{20} \\ \tan^{-1} \frac {x-1}{x+1} & = \tan^{-1} \frac {47}{52} + \tan^{-1} \frac 1{20} \\ & = \tan^{-1} \frac {\frac {47}{52} + \frac 1{20}} {1-\frac {47}{52} \times \frac 1{20}} \\ & = \color{#3D99F6} \tan^{-1} \frac {992}{993} & \small \color{#3D99F6} \text{Multiply up and down by }2 \\ & = \color{#3D99F6} \tan^{-1} \frac {1984}{1986} \\ \therefore \tan^{-1} \frac {{\color{#3D99F6}x}-1}{{\color{#3D99F6}x}+1} & = \tan^{-1} \frac {{\color{#3D99F6}1985}-1}{{\color{#3D99F6}1985}+1} \\ \implies x & = \boxed{1985} \end{aligned}$

Refer to Eli Ross's briiliant solution here to a similar problem to tackle such questions much faster.

I will post a complete solution soon. Here is a brief outline.

Take arctan(1/x) to L.H.S and 3arctan(1/20) to R.H.S.

Take tan on both sides. You get a fraction in L.H.S and use the triple angle formula on R.H.S and simplify.

Then solve for x you get x=1985.