Find the minimum value of f f(x)=tan⁡(x+π6)tan⁡x, x∈(0,π3) My approach is as follows. I tried to solve it by segregating it f(x)=13tan⁡x+(3+13)13−tan⁡x but f′(x) is getting more and more complicated

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Find the minimum value of f  f(x)=tan⁡(x+π6)tan⁡x,    x∈(0,π3)  My approach is as follows. I tried to solve it by segregating it  f(x)=13tan⁡x+(3+13)13−tan⁡x  but f′(x) is getting more and more complicated

Joseph Lewis · Accepted Answer

g(x)=1+12sin⁡xcos⁡(x+π6)=1+1sin⁡(2x+π6)−sin⁡π6  Now we need to maximize   sin⁡(2x+π6)   in   (π6,2π3+π6)

Mason Hall · Accepted Answer

The first derivate of your function is:f′(x)=cot⁡(x)sec⁡(x+π6)2−tan⁡(x+π6)csc⁡(x)2Now, we have to impose f′(x)=0 and so:sin⁡(x)cos⁡(x)−sin⁡(x+π6)cos⁡(x+π6)=0That can be rewritten as:sin⁡(2x)=sin⁡(2x+π3)We know that:sin⁡(α)=sin⁡(β)↔α+β=π+2kπ∨α=β+2kπSo we have:2x=2x+π3+2kπ↔  IMPOSSIBLE∨4x+π3=π+2kπ↔x=π6+kπ2The only solution is x=π6. So:f(π6)=313=3

Find the minimum value of f f(x)=\frac{\tan(x+\frac{\pi}{6})}{\tan x}, \ \ \

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