Question

If $A=\frac{1}{1}{\cot }^{-1}\left(\frac{1}{1}\right)+\frac{1}{2}{\cot }^{-1}\left(\frac{1}{2}\right)+\frac{1}{3}{\cot }^{-1}\left(\frac{1}{3}\right)$ and $B=1{\cot }^{-1}1+2{\cot }^{-1}2+3{\cot }^{-1}3$ then $|B-A|$ is equal to $\frac{a\pi }{b}+\frac{c}{d}{\cot }^{-1}3$ where $a,b,c,d\in N$ and are in their lowest form then $a+b+c+d$ equal to

Answer 1

Given, $A=\frac{1}{1}{\cot }^{-1}\left(\frac{1}{1}\right)+\frac{1}{2}{\cot }^{-1}\left(\frac{1}{2}\right)+\frac{1}{3}{\cot }^{-1}\left(\frac{1}{3}\right)$ and $B=1{\cot }^{-1}1+2{\cot }^{-1}2+3{\cot }^{-1}3$

We know that ${\cot }^{-1}\frac{1}{x}={\tan }^{-1}x\forall x>0$

$\Rightarrow A={\tan }^{-1}1+\frac{1}{2}{\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}3$

Also, ${\tan }^{-1}1={\cot }^{-1}1=\frac{\pi }{4}$

$\therefore |B-A|=|{\cot }^{-1}1+2{\cot }^{-1}2+3{\cot }^{-1}3-({\tan }^{-1}1+\frac{1}{2}{\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}3)|$

$\therefore |B-A|=|2{\cot }^{-1}2+3{\cot }^{-1}3-\frac{1}{2}{\tan }^{-1}2-\frac{1}{3}{\tan }^{-1}3|$

Consider ${\tan }^{-1}2=x$

$\therefore \tan x=2$

$\therefore \sin x=2\sqrt{1-{\sin }^{2}x}$

$\therefore 5{\sin }^{2}x=4$

$\therefore \sin x=\frac{2}{\sqrt{5}}⟹x=\frac{7\pi }{20}$

Also, ${\tan }^{-1}x+{\cot }^{-1}x=\frac{\pi }{2}$

$\therefore |B-A|=|2(\frac{\pi }{2}-\frac{7\pi }{20})+3{\cot }^{-1}3-\frac{7\pi }{40}-\frac{1}{3}(\frac{\pi }{2}-{\cot }^{-1}3)|$

$\therefore |B-A|=|\frac{3\pi }{10}-\frac{7\pi }{40}-\frac{\pi }{6}+3{\cot }^{-1}3+\frac{1}{3}{\cot }^{-1}3|$

$\therefore |B-A|=|-\frac{\pi }{24}+\frac{10}{3}{\cot }^{-1}3|$

$\therefore |B-A|=-\frac{\pi }{24}+\frac{10}{3}{\cot }^{-1}3$ $(\because {\cot }^{-1}3>\frac{\pi }{24})$

$\therefore a+b+c+d=-1+24+10+3=36$