NIMCET 2019 Question Paper with solutions

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Question : 45 of 120

Marks: +1, -0

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

π/6

, then the minimum value of tan A + tan B is

$√3 - √2$
$4- 2√3$
$2/√3$
$2- √3$

Solution:

Given.A > 0, B > 0

A+B=π/6

\tan(A+B)=\;\;{\tan A+\tan B}/{1-\tan A\tan B}

\tan\;\;{π}/{6}=\;\;{\tan A+\tan B}/{1-\tanA\tanB}

1-\tanA\tan B=√{3}(\tan A+\tan B)

\tan A\tan B=1-√{3}(\tan A+\tan B)

By AM ≥ GM

∴\;\;{\tan A+\tan B}/{2}≥√{\tanA\tanB}

\tan A+\tan B≥2√{1-√{3}(\tan A+\tan B)}

⇒(\tan A+\tan B)^{2}≥4(1-√{3}(\tan A+\tan B)

⇒x^{2}+4√{3}x-4≥0[∵\tan A+\tan B=x]

Now,

x^2+4√3x-4=0

⇒x={-4√3±√{48+16}}/2

x={-4√3±8}/2

x=-2√3±4

(x+2√3-4)(x+2√3+4) ≥ 0

Minimum value of

\tan A+\tan B=4-2√3

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