Question

Tan(45+B)=
$2 + \sqrt{3}$
Find B value

Answer 1

tan(45+B)=2+√3

=>tan 45+ tanB/ 1- tan45 tanB = 2+√3

=>1+tanB/1-1×tanB= 2+√3

=>1+tanB/1-tanB=2+√3

=>1+tanB=(2+√3)(1-tanB)

=>1+ tanB=2-2tanB+√3-√3tanB

=>3tanB +√3tanB=2-1+√3

=>tanB(3+√3)=1+√3

=>tanB=1+√3/3+√3

=>tanB=1+√3/3(1+√3)

=>tanB=1/√3

=>tanB=tan30

=>B=30

Answer 2

⛤HERE IS YOUR ANSWER⛤

$tan(45 + \beta ) = 2 + \sqrt{3} \\ or \: \: \frac{tan45 + tan \beta }{1 - tan45.tan \beta } = 2 + \sqrt{3} \\ or \: \: \frac{1 + tan \beta }{1 - tan \beta } = \frac{2 + \sqrt{3} }{1} \\ operting \: we \: get \\ \frac{(1 + tan \beta ) + (1 - tan \beta )}{(1 + tan \beta ) - (1 - tan \beta )} = \frac{(2 + \sqrt{3} ) + 1}{ (2 + \sqrt{3} ) - 1} \\ or \: \: \frac{2}{2tan \beta } = \frac{3 + \sqrt{3} }{1 + \sqrt{3} } \\ or \: \: \frac{1}{tan \beta } = \frac{ \sqrt{3} (1 + \sqrt{3} )}{1 + \sqrt{3} } \\ or \: \: \frac{1}{tan \beta } = \sqrt{3} \\ or \: \: tan \beta = \frac{1}{ \sqrt{3} } = tan \frac{\pi}{6} \\ or \: \: \beta = \frac{\pi}{6}$

⛤HOPE THIS ⬆ HELPS YOU⛤