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

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(D) Given that $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c=\pi$. Using the formula $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c = 	an ^{-1} \left( \frac{a+b+c

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$a+b+c=a b c$

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(D) Given that $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c=\pi$. Using the formula $	an ^{-1} a+	an ^{-1} b+	an ^{-1} c = 	an ^{-1} \left( \frac{a+b+c-abc}{1-ab-bc-ca} ight) = \pi$. Taking $	an$ on both sides,we get $\frac{a+b+c-abc}{1-ab-bc-ca} = 	an \pi$. Since $	an \pi = 0$,we have $\frac{a+b+c-abc}{1-ab-bc-ca} = 0$. This implies $a+b+c-abc = 0$. Therefore,$a+b+c=abc$.

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

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