If tanα = \(\frac{1}{2}\), tanβ = \(\frac{1}{3}\), then find the value of [tan(2α+β)+2].

1 3 5 7

5

[Note: If tanα =\(\frac{1}{2}\), tanβ=\(\frac{1}{3}\) then tan(α+β)=45°] Now, ⇒ tan(2α+β)+2 = tan(α + α + β) + 2 = tan (α + 45°) + 2 = \(\frac{tanα\;+\;tan45°}{1\;-\;tanα\;tan45°}\) + 2 =\(\frac{\frac{1}{2}+1}{1-\frac{1}{2}}\) + 2 =\(\frac{3}{2}\) × \(\frac{2}{1}\) + 2 = 5