The simplified form of tan^{-1}left(frac{x}{y | vedclass

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Question

(B) Given expression: $	an^{-1}\left(\frac{x}{y}ight) - 	an^{-1}\left(\frac{x-y}{x+y}ight)$.Divide the numerator and denominator of the second t

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Given expression: $	an^{-1}\left(\frac{x}{y}ight) - 	an^{-1}\left(\frac{x-y}{x+y}ight)$.Divide the numerator and denominator of the second term by $x$:$	an^{-1}\left(\frac{x}{y}ight) - 	an^{-1}\left(\frac{1 - \frac{y}{x}}{1 + \frac{y}{x}}ight)$.Using the identity $	an^{-1}(A) - 	an^{-1}(B) = 	an^{-1}\left(\frac{A-B}{1+AB}ight)$,we have:$	an^{-1}\left(\frac{1 - \frac{y}{x}}{1 + 1 \cdot \frac{y}{x}}ight) = 	an^{-1}(1) - 	an^{-1}\left(\frac{y}{x}ight)$.Substituting this back into the expression:$= 	an^{-1}\left(\frac{x}{y}ight) - \left(	an^{-1}(1) - 	an^{-1}\left(\frac{y}{x}ight)ight)$$= 	an^{-1}\left(\frac{x}{y}ight) + 	an^{-1}\left(\frac{y}{x}ight) - 	an^{-1}(1)$.Since $	an^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$ and $\cot^{-1}(z) = 	an^{-1}\left(\frac{1}{z}ight)$:$= 	an^{-1}\left(\frac{x}{y}ight) + \cot^{-1}\left(\frac{x}{y}ight) - 	an^{-1}(1)$$= \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

The simplified form of $\tan^{-1}\left(\frac{x}{y}\right) - \tan^{-1}\left(\frac{x-y}{x+y}\right)$ is equal to

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