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

Correct Answer: 47

Given:

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

Combine with \(\cot^{-1} 5\):

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

Simplify the expression:

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

Add \(\cot^{-1} n\):

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

Using the identity:

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

the equation becomes:

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

Further simplification yields:

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

Apply the tangent addition formula:

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

This leads to:

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

Simplify numerator and denominator:

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

Cancel \(23n\) and solve: \[ \frac{24n + 23}{n - 24} = 1. \]

Cross-multiplication gives: \[ 24n + 23 = n - 24. \]

Solving for n: \[ 23n = 47. \]

Therefore: \[ n = 47. \]