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If tan^(-1)a+tan^(-1)b+tan^(-1)c=pi then...

If `tan^(-1)a+tan^(-1)b+tan^(-1)c=pi` then prove tjhat `a+b+c=abc`

A

`a+b-c=(ab)/(c)`

B

`a+b+c=abc`

C

`a+b+c=1`

D

abc =1

Text Solution

Verified by Experts

The correct Answer is:
B
Given , `tan^(-1)a+tan^(-1)b+tan^(-1)x=pi`
`rArrtan^(-1)((a+b)/(1-ab))+tan^(-1)x=pi`
`rArrtan^(-1)(((a+b)/(1-ab)+c)/(1-(a+b)/(1-ab)))=pi`
`rArr((a+b)/(1-ab)+c)/(1-(a+b)/(1-ab)*c)=tan^(-1)pi=0`
`rArra+b+c-abc=0`
`rArra+b+c=abc`
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