Question

if tan ( a+b ) =√3 and tan a= 1/√3 then tan b =

(a) 1/√3
(b) 0
(c) 1
(d) √3

Answer 1

$\tan(a + b) = \sqrt{3} \\ \frac{ \tan(a )+ \tan(b) }{ 1 - \tan(a) \tan(b )} = \sqrt{3} \\ \tan(a) + \tan(b) = \sqrt{3} - \sqrt{3} \tan(a) \tan(b) \\ \tan(b) = \frac{ \sqrt{3} - \tan(a) }{1 + \sqrt{3} \tan(a) } \\ = \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} \times \frac{1}{ \sqrt{3} } } = \frac{ \frac{2}{ \sqrt{3} } }{2} = \frac{1}{ \sqrt{3} }$
Option (a)

Answer 2

●WE KNOW THAT :-

$= > tan(a + b) = \frac{tana + tanb}{1 - tana.tanb} \: \: \: ..[Eq _{1}]$
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◢GIVEN

$= > tan(a + b) = \sqrt{3} \\ \\ and \\ \\ = > tana = \frac{1}{ \sqrt{3} }$
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●PUT THE VALUE IN Eq(1)

$= > \sqrt{3} = \frac{ \frac{1}{ \sqrt{3} } + tanb }{1 - \frac{1}{ \sqrt{3} } tanb} \\ \\ = > \sqrt{3} - \sqrt{3} \times \frac{1}{ \sqrt{3} } .tanb = \frac{1}{ \sqrt{3} } + tanb \\ \\ = > \sqrt{3} - tanb = \frac{1}{ \sqrt{3} } + tanb \\ \\ = > 2tanb = \sqrt{3} - \frac{1}{ \sqrt{3} } \\ \\ = > 2tanb = \frac{3 - 1}{ \sqrt{3} } = \frac{2}{ \sqrt{3} } \\ \\ = > tanb = \frac{1}{ \sqrt{3} } \: \: \: \: \: \: ...[ANSWER]$
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●HENCE THE OPTION "a" IS THE CORRECT ANSWER.

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