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

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: \[ n = 47. \]

Answer 2

We are asked to find the value of \( n \in \mathbb{N} \) that satisfies the equation \( \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} \).

Concept Used:

To solve this equation involving inverse trigonometric functions, we use the following key identities:

1. The relationship between \( \cot^{-1} x \) and \( \tan^{-1} x \):

\[ \cot^{-1} x = \tan^{-1} \left(\frac{1}{x}\right) \quad \text{for } x > 0 \]

2. The addition formula for \( \tan^{-1} x \):

\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left(\frac{a+b}{1-ab}\right) \quad \text{for } ab < 1 \]

3. The subtraction formula for \( \tan^{-1} x \):

\[ \tan^{-1} a - \tan^{-1} b = \tan^{-1} \left(\frac{a-b}{1+ab}\right) \quad \text{for } ab > -1 \]

We also use the standard value \( \tan^{-1}(1) = \frac{\pi}{4} \).

Step-by-Step Solution:

Step 1: Convert all the \( \cot^{-1} \) terms into \( \tan^{-1} \) terms using the identity \( \cot^{-1} x = \tan^{-1}(1/x) \).

The given equation is:

\[ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} \]

Converting to \( \tan^{-1} \):

\[ \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{n}\right) = \frac{\pi}{4} \]

Step 2: Combine the first two terms, \( \tan^{-1}(1/3) + \tan^{-1}(1/4) \), using the addition formula.

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

The equation now becomes:

\[ \tan^{-1}\left(\frac{7}{11}\right) + \tan^{-1}\left(\frac{1}{5}\right) + \tan^{-1}\left(\frac{1}{n}\right) = \frac{\pi}{4} \]

Step 3: Combine the result from Step 2 with the third term, \( \tan^{-1}(1/5) \).

\[ \tan^{-1}\left(\frac{7}{11}\right) + \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{7}{11} + \frac{1}{5}}{1 - \frac{7}{11} \cdot \frac{1}{5}}\right) = \tan^{-1}\left(\frac{\frac{35+11}{55}}{1 - \frac{7}{55}}\right) = \tan^{-1}\left(\frac{\frac{46}{55}}{\frac{48}{55}}\right) = \tan^{-1}\left(\frac{46}{48}\right) = \tan^{-1}\left(\frac{23}{24}\right) \]

The equation is now simplified to:

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

Step 4: Rearrange the equation and use the value \( \frac{\pi}{4} = \tan^{-1}(1) \).

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

Final Computation & Result:

Step 5: Apply the subtraction formula for \( \tan^{-1} \) to the right side of the equation.

\[ \tan^{-1}\left(\frac{1}{n}\right) = \tan^{-1}\left(\frac{1 - \frac{23}{24}}{1 + 1 \cdot \frac{23}{24}}\right) \] \[ \tan^{-1}\left(\frac{1}{n}\right) = \tan^{-1}\left(\frac{\frac{24-23}{24}}{\frac{24+23}{24}}\right) = \tan^{-1}\left(\frac{\frac{1}{24}}{\frac{47}{24}}\right) = \tan^{-1}\left(\frac{1}{47}\right) \]

Step 6: Equate the arguments of the \( \tan^{-1} \) function to solve for \( n \).

\[ \frac{1}{n} = \frac{1}{47} \] \[ n = 47 \]

The value of \( n \) is 47.