Evaluate: tan^{-1} left( frac{a - b}{1 + ab} | vedclass

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(B) We use the standard identity for the difference of inverse tangents: $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.A

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$	an^{-1} a - 	an^{-1} c$

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(B) We use the standard identity for the difference of inverse tangents: $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$.Applying this to the given expression:$	an^{-1} \left( \frac{a - b}{1 + ab} ight) = 	an^{-1} a - 	an^{-1} b$Similarly:$	an^{-1} \left( \frac{b - c}{1 + bc} ight) = 	an^{-1} b - 	an^{-1} c$Adding these two results together:$(	an^{-1} a - 	an^{-1} b) + (	an^{-1} b - 	an^{-1} c)$The term $-	an^{-1} b$ and $+	an^{-1} b$ cancel each other out:$= 	an^{-1} a - 	an^{-1} c$.

Evaluate: $\tan^{-1} \left( \frac{a - b}{1 + ab} \right) + \tan^{-1} \left( \frac{b - c}{1 + bc} \right)$

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