Arc Again And Tangent Again!

Relevant wiki: Complex Plane

Note that $\tan^{-1}\frac{1}{n}$ is the argument of $n+i,$ and adding angles is essentially like multiplying the arguments. Note that $(3+i)(4+i)(5+i)(n+i) = (48n-46) + (48+46n)i.$

Since the RHS is $\frac{\pi}{4},$ the real and imaginary parts of the result above must be equal. Setting $48n-46 = 48+46n$ gives $2n = 94,$ so $n=47.$

See this for a quick review of Inverse-Trigonometric-Functions.

$\color{#20A900}{\arctan \dfrac{1}{3} + \arctan \dfrac{1}{4}+\arctan \dfrac{1}{5}}= \dfrac{\pi}{4}-\arctan \dfrac{1}{n}$ Taking $\tan$ on both sides and using:- $\color{#D61F06}{\tan(A+B+C)=\dfrac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan B\tan C -\tan C\tan A}}\\\\ \color{#EC7300}{\arctan(\tan A)=A} \text{ and } \color{#3D99F6}{ \tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}}$

(For the above identities see Tangent of Sums here )

$\dfrac{\dfrac 13 +\dfrac 14 +\dfrac 15-\dfrac{1}{3\cdot 4\cdot 5}}{1-\dfrac{1}{3\cdot4}-\dfrac{1}{4\cdot 5}-\dfrac{1}{5\cdot3}}=\dfrac{1-\dfrac{1}{n}}{1+\dfrac{1}{n}}$ $\dfrac{46}{48}=\dfrac{n-1}{n+1}$ $\dfrac{47-1}{47+1}=\dfrac{n-1}{n+1}$ $\Huge \therefore \color{#456461}{\boxed{\color{#EC7300}{\boxed{\color{#0C6AC7}{n=47}}}}}$

Since we are dealing with acute angles, $\tan(\arctan a) = a$

Note that, $\tan(\arctan a + \arctan b)= \dfrac{a+b}{1-ab}$ , by tangent addition. Thus, $\arctan a + \arctan b = \arctan \dfrac{a+b}{1-ab}$

Applying this to first two terms, we get $\arctan \dfrac{1}{3}+ \arctan \dfrac{1}{4} = \arctan \dfrac{7}{11}$ .

Now, $\arctan \dfrac{7}{11} + \arctan \dfrac{1}{5} = \arctan \dfrac{23}{24}$ .

We now have $\arctan \dfrac{23}{24}+ \arctan \dfrac{1}{n} = \dfrac{\pi}{4} = \arctan 1$ .

Thus, $\dfrac{\dfrac{23}{24} + \dfrac{1}{n}}{1- \dfrac{23}{24n}}$ and simplifying ;

$23n +24= 24n - 23 \to n = 47$