The value of tan ^{-1}left(frac{x}{y}right)-t | vedclass

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Question

(A) We use the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A-B}{1+AB} ight)$.Let $A = \frac{x}{y}$ and $B = \frac{x-y}{x+y}$.Then

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$\frac{\pi}{4}$

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(A) We use the formula $	an ^{-1} A - 	an ^{-1} B = 	an ^{-1} \left( \frac{A-B}{1+AB} ight)$.Let $A = \frac{x}{y}$ and $B = \frac{x-y}{x+y}$.Then,$	an ^{-1} \left( \frac{x}{y} ight) - 	an ^{-1} \left( \frac{x-y}{x+y} ight) = 	an ^{-1} \left( \frac{\frac{x}{y} - \frac{x-y}{x+y}}{1 + \frac{x}{y} \cdot \frac{x-y}{x+y}} ight)$.Simplifying the numerator: $\frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2 + xy - xy + y^2}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$.Simplifying the denominator: $1 + \frac{x^2 - xy}{y(x+y)} = \frac{xy + y^2 + x^2 - xy}{y(x+y)} = \frac{x^2 + y^2}{y(x+y)}$.Thus,the expression becomes $	an ^{-1} \left( \frac{\frac{x^2 + y^2}{y(x+y)}}{\frac{x^2 + y^2}{y(x+y)}} ight) = 	an ^{-1}(1)$.Since $	an ^{-1}(1) = \frac{\pi}{4}$,the final value is $\frac{\pi}{4}$.

The value of $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1}\left(\frac{x-y}{x+y}\right)$ (where,$x, y>0$) is

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