For n N, if ^-13+ ^-14+ ^-15+ ^-1n=( )/(4), then n is equal to ________.

Question

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

Answer: 47

Accepted Answer

For $n\in N$, 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:

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

Now, adding the third term:

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

we apply the identity again:

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

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

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

This simplifies to:

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

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:

$n=47$

Solutions · 1