Question

For $n \in \mathbb{N}$, if $\cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4},$ then $n$ is equal to ______.

Answer 1

We know:

$\cot^{-1} 3 + \cot^{-1} 4 = \tan^{-1} \left( \frac{3 \times 4 - 1}{3 + 4} \right) = \tan^{-1} \left( \frac{12 - 1}{7} \right) = \tan^{-1} \left( \frac{11}{7} \right).$

Adding $\cot^{-1} 5$:

$\tan^{-1} \left( \frac{11}{7} \right) + \cot^{-1} 5 = \tan^{-1} \left( \frac{\frac{11}{7} \times 5 - 1}{\frac{11}{7} + 5} \right) = \tan^{-1} \left( \frac{\frac{55}{7} - 1}{\frac{11}{7} + 5} \right).$

Simplify:

$= \tan^{-1} \left( \frac{48}{46} \right) = \tan^{-1} \left( \frac{24}{23} \right).$

Adding $\cot^{-1} n$:

$\tan^{-1} \left( \frac{24}{23} \right) + \cot^{-1} n = \frac{\pi}{4}.$

Using the identity:

$\cot^{-1} a + \cot^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right),$

we rewrite:

$\tan^{-1} \left( \frac{24}{23} \right) + \cot^{-1} n = \frac{\pi}{4}.$

Simplify further:

$\tan^{-1} \left( \frac{24}{23} \right) + \tan^{-1} \left( \frac{1}{n} \right) = \frac{\pi}{4}.$

Using the tangent addition formula:

$\tan \left( \tan^{-1} \left( \frac{24}{23} \right) + \tan^{-1} \left( \frac{1}{n} \right) \right) = 1.$

This implies:

$\frac{\frac{24}{23} + \frac{1}{n}}{1 - \frac{24}{23} \times \frac{1}{n}} = 1.$

Simplify the numerator and denominator:

$\frac{\frac{24n + 23}{23n}}{\frac{n - 24}{23n}} = 1.$

Cancel $23n$ and solve: $\frac{24n + 23}{n - 24} = 1.$

Cross-multiply: $24n + 23 = n - 24.$

Simplify: $23n = 47.$

Thus: $n = 47.$