QJEEM24Macb1d: JEE Main 2024 Mathematics - Class 12 JEEM24Macb1d Solution: For , if , then is equal to ________.

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 .

Given the equation:

$\cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4}$

we can use the identity for the sum of inverse cotangents. Starting with the first two terms:

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

Now, adding the third term:

$\cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}(5)$

we apply the identity again:

$\cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}\left(\frac{n \times 5 - 1}{5 + n}\right) = \frac{\pi}{4}$

Rewriting this to isolate the sum of the terms, we proceed as follows:

$\cot^{-1}\left( \frac{\left(\frac{11}{7} \times \frac{5n-1}{5+n} - 1\right)}{\left(\frac{11}{7} + \frac{5n-1}{5+n} \right)} \right) = \frac{\pi}{4}$

This simplifies to:

$\frac{11}{7} \left(\frac{5n-1}{5+n}\right) - 1 = \frac{11}{7} + \frac{5n-1}{5+n}$

Solving the equation:

$\frac{55n - 11}{5 + n} - 1 = \frac{11}{7} + \frac{5n - 1}{5 + n}$

Further simplification yields:

$55n - 11 - 35 - 7n = 55 + 11n + 35n - 7$

Bringing the terms together, we get:

$48n - 46 = 48$

Therefore:

$2n = 94$

So finally:

$n = 47$