JEE Main 2016 · 10 Apr · Shift 1 · Mathematics Q88

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MathematicsSingle correct+4 / -1Question 88 of 90

If A > 0, B > 0 and $A+B=\dfrac\pi6$ , then the minimum value of tanA + tanB is :

\sqrt3-\sqrt2

2-\sqrt3

4-2\sqrt3

\dfrac2{\sqrt3}

Answer and worked solution

Official answer: C

$\tan A+\tan B=\frac{\sin(A+B)}{\cos A\cos B}=\frac{1/2}{\cos A\cos B}$ , smallest when $\cos A\cos B$ is largest, i.e. A = B = 15°: $2\tan15^\circ=4-2\sqrt3$ .

Topics:Trigonometric ratios and identities Maxima and minima

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JEE Main 2016 · 10 Apr · Shift 1, Mathematics question 88