Evaluate: cot ^{ - 1}left(frac{xy + 1}{x - y} | vedclass

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(A) We know the identity for the difference of inverse cotangent functions: $\cot ^{ - 1}\left(\frac{ab + 1}{a - b}\right) = \cot ^{ - 1}b - \cot ^{ -

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) We know the identity for the difference of inverse cotangent functions: $\cot ^{ - 1}\left(\frac{ab + 1}{a - b}\right) = \cot ^{ - 1}b - \cot ^{ - 1}a$. Applying this identity to each term in the expression: $\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) = \cot ^{ - 1}y - \cot ^{ - 1}x$ $\cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) = \cot ^{ - 1}z - \cot ^{ - 1}y$ $\cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right) = \cot ^{ - 1}x - \cot ^{ - 1}z$ Adding these three expressions together: $(\cot ^{ - 1}y - \cot ^{ - 1}x) + (\cot ^{ - 1}z - \cot ^{ - 1}y) + (\cot ^{ - 1}x - \cot ^{ - 1}z)$ Canceling the terms: $= (\cot ^{ - 1}y - \cot ^{ - 1}y) + (\cot ^{ - 1}z - \cot ^{ - 1}z) + (\cot ^{ - 1}x - \cot ^{ - 1}x) = 0$ Thus,the correct option is $A$.

Evaluate: $\cot ^{ - 1}\left(\frac{xy + 1}{x - y}\right) + \cot ^{ - 1}\left(\frac{yz + 1}{y - z}\right) + \cot ^{ - 1}\left(\frac{zx + 1}{z - x}\right)$

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