Question

Let $f(x)={\tan }^{-1}(\cot x-2\cot 2x)$ and $\sum _{r=1}^{5}f(r)=a-b\pi$, where $a,b\in N$. Find the value of $(a+b)$

Answer 1

Answer: 20

Given, $f(x)={\tan }^{-1}(\cot x-2\cot 2x)$
$={\tan }^{-1}\left(\frac{1}{\tan x}-\frac{1-{\tan }^{2}x}{\tan x}\right)={\tan }^{-1}\left(\frac{{\tan }^{2}x}{\tan x}\right)={\tan }^{-1}(\tan x).$
Now, $\sum _{r=1}^{5}f(r)=f(1)+f(2)+f(3)+f(4)+f(5)$
As, $f(1)=1,f(2)=2-\pi ,f(3)=3-\pi ,f(4)=4-\pi ,f(5)=5-2\pi$
$\Rightarrow \sum _{r=1}^{5}f(r)=15-5\pi =a-b\pi \Rightarrow a=15,b=5$
Hence, the value of $(a+b)=20$.