If tan ^{-1} a+tan ^{-1} b+tan ^{-1} c=pi,the | vedclass

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(C) We are given the equation: $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c=\pi$. Using the formula for the sum of three inverse tangent functions: $	an ^

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$a+b+c=a b c$

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(C) We are given the equation: $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c=\pi$. Using the formula for the sum of three inverse tangent functions: $	an ^{-1} x + 	an ^{-1} y + 	an ^{-1} z = 	an ^{-1} \left( \frac{x+y+z-xyz}{1-xy-yz-zx} ight)$. Applying this to the given equation: $	an ^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} ight) = \pi$. Taking the tangent of both sides: $\frac{a+b+c-abc}{1-ab-bc-ca} = 	an \pi$. Since $	an \pi = 0$,we have: $\frac{a+b+c-abc}{1-ab-bc-ca} = 0$. This implies that the numerator must be zero: $a+b+c-abc = 0$. Therefore,$a+b+c = abc$.

If $\tan ^{-1} a+\tan ^{-1} b+\tan ^{-1} c=\pi$,then which of the following is true?

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