If A > 0, B > 0 and A + B = π/6 , then the minimum value of tanA + tanB

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Question Id : 3904 | Context :NIMCET 2019

If A > 0, B > 0 and A + B =

$\frac{\pi}{6}$ , then the minimum value of

$tanA + tanB$

🎥 Video solution / Text Solution of this question is given below:

On differentiating

x= tanA + tan(π/6-A)

we get :

dx/dA = sec²A-sec²(π/6-A)

now putting

dx/dA=0

we get

cos²(A) = cos²(π/6-A) so 0≤A≤π/6

therefore

A=π/6-A from here we get A = π/12 = B

so minimum value of that function is

2tanπ/12 which is equal to 2(2-√3)

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