JEE Main 2024 — Inverse Trigonometric Functions Question with Solution

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:

$n=47$