Question

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

Answer 1

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