Question

$co{t}^{-1}\left(\frac{xy+1}{x-y}\right)+co{t}^{-1}\left(\frac{yz+1}{y-z}\right)+co{t}^{-1}\left(\frac{zx+1}{z-x}\right)=?$

• A. $0$
• B. $1$
• C. $co{t}^{-1}\left(x\right)+co{t}^{-1}\left(y\right)+co{t}^{-1}\left(z\right)$
• D. None of these

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Given, $co{t}^{-1}\left(\frac{xy+1}{x-y}\right)+co{t}^{-1}\left(\frac{yz+1}{y-z}\right)+co{t}^{-1}\left(\frac{zx+1}{z-x}\right)$

$={\tan }^{-1}\left(\frac{x-y}{xy+1}\right)+{\tan }^{-1}\left(\frac{y-z}{yz+1}\right)+{\tan }^{-1}\left(\frac{z-x}{zx+1}\right)$

Use the identity $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathbf{-}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{B}\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\frac{\mathbf{A}\mathbf{-}\mathbf{B}}{\mathbf{1}\mathbf{+}\mathbf{A}\mathbf{\times }\mathbf{B}}\mathbf{\right)}$

$=\left[{\tan }^{-1}\right(x)–{\tan }^{-1}(y\left)\right]+\left[{\tan }^{-1}\right(y)–{\tan }^{-1}(z\left)\right]+\left[{\tan }^{-1}\right(z)–{\tan }^{-1}(x\left)\right]$

$=\mathbf{0}$

Hence, option ‘A’ is correct.