Question

${\cot }^{-1}\frac{xy+1}{x-y}+{\cot }^1\frac{yz+1}{y-z}+{\cot }^{-1}\frac{xz+1}{z-x}$

• A. 1
• B. -1
• C. 0
• D. none of these

Answer 1

The correct option is C 0

Take $x=\tan x,y=\tan B,z=\tan r$

Then

${\cot }^{-1}\frac{xy+1}{x-y}={\cot }^{-1}\left(\frac{\tan \alpha \tan \beta +1}{\tan \alpha -\tan \beta }\right)$

$={\cot }^{-1}\frac{1}{\tan (\alpha -\beta )}={\cot }^{-1}\cot (\alpha -\beta )$

$\alpha -\beta$

Similarly ${\cot }^{-1}\frac{yz+1}{y-z}=\beta -r$

${\cot }^{-1}\frac{zx+1}{z-x}=r=\alpha$

$\therefore {\cot }^{-1}\frac{xy+1}{x-y}+{\cot }^{-1}\frac{yz+1}{y-z}+{\cot }^{-1}\frac{zx+1}{z-x}$

$=\alpha -\beta +\beta -r+r-\alpha$

$=0$