If A > 0, B > 0 and `A+B=pi/6`, then the minimum value of `tan A + tan B` is

Options

A. `2-sqrt3`
B. `(2)/(sqrt3)`
C. `sqrt3-sqrt2`
D. `4-2sqrt3`

Answer

D

Explanation

`tan A +tan D`
`=tanA+tan((pi)/(6)-A)=tanA+(tan""(pi)/(6)-tanA)/(1+tan""(pi)/(6)tanA)`
`=tanA+(1-sqrt3tanA)/(sqrt3+tanA)=(1+tan^(2)A)/(sqrt3+tanA)=(1)/(sqrt3cos^(2)A+sinA cos A)`
`=(2)/(sqrt3(1+cos2A)+sin2A)=(2)/(sqrt3+(sqrt3cos2A+sin2A))`
`=(1)/((sqrt3)/(2)+sin(2A+(pi)/(3))`
`Now, 0ltAlt(pi)/(6)implies0lt2Alt(pi)/(3)`
`implies(pi)/(3)lt2A+(pi)/(3)lt(2pi)/(3)implies(sqrt3)/(2)ltsin(2A+(pi)/(3))le1`
`impliessqrt3lt(sqrt3)/(2)+sin(2A+(pi)/(3))le(sqrt3)/(2)+1`
`implies(2)/(2+sqrt3)le(1)/((sqrt3)/(2)+sin(2A+(pi)/(3)))lt(1)/(sqrt3)`
`implies2(2-sqrt3)le tanA+tanBlt(1)/(sqrt3)` lt