JEE Main 2016 — Mathematics Trigonometry

$$\mathrm{tan}⁡(A+B)=\frac{\mathrm{tan}⁡A+\mathrm{tan}⁡B}{1-\mathrm{tan}⁡A\mathrm{tan}⁡B}$$tan⁡(A+B)=1−tan⁡Atan⁡Btan⁡A+tan⁡B​

$\Rightarrow$⇒ $\frac{1}{\sqrt{3}}=\frac{y}{1-\mathrm{tan}⁡A \mathrm{tan}⁡B}$3​1​=1−tan⁡Atan⁡By​ where $y=\mathrm{tan}⁡A+\mathrm{tan}⁡B$y=tan⁡A+tan⁡B

$\Rightarrow$⇒ $\mathrm{tan}⁡A.\mathrm{tan}⁡B=1-\sqrt{3}y$tan⁡A.tan⁡B=1−3​y

and, also $\mathrm{AM}\geq \mathrm{GM}$AM≥GM

$\Rightarrow$⇒ $\frac{\mathrm{tan}⁡A+\mathrm{tan}⁡B}{2} \geq \sqrt{\mathrm{tan}⁡A\mathrm{tan}⁡B}$2tan⁡A+tan⁡B​≥tan⁡Atan⁡B​

$\Rightarrow$⇒ $y \geq 2\sqrt{1- \sqrt{3}y}$y≥21−3​y​

$\Rightarrow$⇒ ${y}^{2} \geq 4-4\sqrt{3}y\geq 0$y2≥4−43​y≥0

$\Rightarrow$⇒ $y \leq -2\sqrt{3}-4$y≤−23​−4 or $y \geq -2\sqrt{3}+4$y≥−23​+4

$y \leq -2\sqrt{3}-4$y≤−23​−4, is not possible as $\mathrm{tan}⁡A+\mathrm{tan}⁡B>0$tan⁡A+tan⁡B>0

Therefore, $y \geq -2\sqrt{3}+4$y≥−23​+4

Hence, minimum positive value is $4-2\sqrt{3}$4−23​.