Question

if a b c are distinct non zero real numbers having the same sign prove that cot inverse AB + 1 by a + b + cot inverse B C + 1 by b minus C + cot inverse CA + 1 by C -a = phi or 2phi​

Answer 1

To prove that, $\cot^{-1}\dfrac{ab+1}{a-b} +\cot^{-1}\dfrac{bc+1}{b-c} +\cot^{-1}\dfrac{ca+1}{a-b}$ = 0.

Solution:

L.H.S. = $\cot^{-1}\dfrac{ab+1}{a-b} +\cot^{-1}\dfrac{bc+1}{b-c} +\cot^{-1}\dfrac{ca+1}{a-b}$

Using the inverse trigonometric identity,

$\cot^{-1} x$ =  $\tan^{-1}\dfrac{1}{x}$

= $\tan^{-1}\dfrac{a-b}{ab+1} +\tan^{-1}\dfrac{b-c}{bc+1} } +\tan^{-1}\dfrac{c-a}{ca+1}$

Using the inverse trigonometric identity,

$\tan^{-1} x + \tan^{-1} y$ = $\tan^{-1}\dfrac{x-y}{xy+1}$

= ($\tan^{-1} a$ - $\tan^{-1} b$) + ($\tan^{-1} b$ - $\tan^{-1} c$ ) + ($\tan^{-1} c$ - $\tan^{-1} a$)

= $\tan^{-1} a$ - $\tan^{-1} b$ + $\tan^{-1} b$ - $\tan^{-1} c$ + $\tan^{-1} c$ - $\tan^{-1} a$

= 0

= R.H.S., proved.

Thus, $\cot^{-1}\dfrac{ab+1}{a-b} +\cot^{-1}\dfrac{bc+1}{b-c} +\cot^{-1}\dfrac{ca+1}{a-b}$ = 0, proved.

Answer 2

$\underline {\green {Solution :}}$

Since , (a-b)+(b-c)+ (c-a) = 0, and (a-b),(b-c) and (c-a) all cannot have the same sign .

Now two cases will rise namely , either two of these numbers are positive and one negative or two of these numbers are negative and one positive.

$\underline {\green {case 1 :}}$

we assume that

(a-b), (b-c) are both positive and (c-a) is negative.

$cot^{-1} \big( \frac{ab+1}{a-b} \big) \\=tan^{-1} \big( \frac{a-b}{1+ab} \big) \\= tan^{-1} a - tan^{-1} b \: ( Since , \:ab > 0 )$

$cot^{-1} \big( \frac{bc+1}{b-c} \big) \\=tan^{-1} \big( \frac{b-c}{1+bc} \big) \\= tan^{-1} b - tan^{-1} c \: ( Since ,\: cb > 0 )$

$and \: cot^{-1} \big( \frac{ca+1}{c-a} \big) = \pi + tan^{-1} \big( \frac{c-a}{1+ca} \big) \\= \pi + ( tan^{-1} c - tan^{-1} a ) \: ( since, \: ca > 0 )$

$\therefore \Sum cot^{-1} \big( \frac{ab+1}{a-b} \big) \\= ( tan^{-1} a - tan^{-1} b) +( tan^{-1} b - tan^{-1} c) + \pi + ( tan^{-1} c - tan^{-1} a ) = \pi$

$\underline {\green {case 2 :}}$

we assume that (a-b) and (b-c) are both negative and (c-a) is positive.

$cot^{-1} \big( \frac{ab+1}{a-b} \big) \\ = \pi - cot^{-1} \big( \frac{ab+1}{b-a} \big) \\= \pi - tan^{-1} \big( \frac{b-a}{1+ab} \big) \\ = \pi - ( tan^{-1} b - tan^{-1} a )$

$cot^{-1} \big( \frac{bc+1}{b-c} \big) \\ = \pi - cot^{-1} \big( \frac{bc+1}{c-b} \big) \\= \pi - tan^{-1} \big( \frac{c-b}{1+bc} \big) \\ = \pi - ( tan^{-1} c - tan^{-1} b )$

$cot^{-1} \big( \frac{ca+1}{c-a} \big) \\ = tan^{-1} \big( \frac{c-a}{1+ca} \big) \\ = tan^{-1} c - tan^{-1} a$

$\therefore \sum cot^{-1} \big( \frac{ab+1}{a-b} \big) \\= \pi - ( tan^{-1} b - tan^{-1} a) + \pi - ( tan^{-1} c - tan^{-1} b) + ( tan^{-1} c - tan^{-1} a ) =2 \pi$

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