Question

If a > b > c > 0 then prove that cot⁻¹(ab+1/a-b)+cot⁻¹(bc+1/b-c)+cot⁻¹(ca+1/c-a)=π

Answer 1

Answer:

$\bf\:cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+cot^{-1}(\frac{ca+1}{c-a})=\pi$

Step-by-step explanation:

If a > b > c > 0 then prove that cot⁻¹(ab+1/a-b)+cot⁻¹(bc+1/b-c)+cot⁻¹(ca+1/c-a)=π

$cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+cot^{-1}(\frac{ca+1}{c-a})$

$\text{since c-a is negative c-a = -(a-c)}$

$=cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+cot^{-1}(\frac{ac+1}{-(a-c)})$

using

$\bf\boxed{cot^{-1}(-x)=\pi-cot^{-1}x}$

$=cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+\pi-cot^{-1}(\frac{ac+1}{a-c})$

using

$\bf\boxed{cot^{-1}x=tan^{-1}\frac{1}{x}}$

$=tan^{-1}(\frac{a-b}{ab+1})+tan^{-1}(\frac{b-c}{bc+1})+\pi-tan^{-1}(\frac{a-c}{ac+1})$

using

$\bf\boxed{\tan^{-1}x-tan^{-1}y=tan^{-1}(\frac{x-y}{xy+1})}$

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

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

$=\pi$

$\implies\bf\:cot^{-1}(\frac{ab+1}{a-b})+cot^{-1}(\frac{bc+1}{b-c})+cot^{-1}(\frac{ca+1}{c-a})=\pi$