Solve tan ^{-1}left(frac{x}{y}right)-tan ^{-1 | 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 $B = \frac{x-y}{x+y}$.$	a

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(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 $B = \frac{x-y}{x+y}$.$	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+\left(\frac{x}{y}ight)\left(\frac{x-y}{x+y}ight)}ight]$$= 	an ^{-1}\left[\frac{\frac{x(x+y)-y(x-y)}{y(x+y)}}{\frac{y(x+y)+x(x-y)}{y(x+y)}}ight]$$= 	an ^{-1}\left(\frac{x^{2}+xy-xy+y^{2}}{xy+y^{2}+x^{2}-xy}ight)$$= 	an ^{-1}\left(\frac{x^{2}+y^{2}}{x^{2}+y^{2}}ight) = 	an ^{-1}(1) = \frac{\pi}{4}$.

Solve $\tan ^{-1}\left(\frac{x}{y}\right)-\tan ^{-1} \left(\frac{x-y}{x+y}\right)$ is equal to

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