tan^{-1}(-2) - tan^{-1}(3) is equal to | vedclass

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(D) Given expression: $	an^{-1}(-2) - 	an^{-1}(3)$Using the property $	an^{-1}(-x) = -	an^{-1}x$,we get:$= -	an^{-1}(2) - 	an^{-1}(3)$$= -(	an^

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$\frac{-3 \pi}{4}$

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(D) Given expression: $	an^{-1}(-2) - 	an^{-1}(3)$Using the property $	an^{-1}(-x) = -	an^{-1}x$,we get:$= -	an^{-1}(2) - 	an^{-1}(3)$$= -(	an^{-1}(2) + 	an^{-1}(3))$Since $xy = 2 	imes 3 = 6 > 1$,we use the formula $	an^{-1}x + 	an^{-1}y = \pi + 	an^{-1}\left(\frac{x + y}{1 - xy}ight)$:$= -\left(\pi + 	an^{-1}\left(\frac{2 + 3}{1 - 2 	imes 3}ight)ight)$$= -\pi - 	an^{-1}\left(\frac{5}{1 - 6}ight)$$= -\pi - 	an^{-1}(-1)$Since $	an^{-1}(-1) = -\frac{\pi}{4}$,we have:$= -\pi - (-\frac{\pi}{4})$$= -\pi + \frac{\pi}{4}$$= -\frac{3 \pi}{4}$

$\tan^{-1}(-2) - \tan^{-1}(3)$ is equal to

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