JEE Main 2023 — Mathematics Trigonometry

Given,

${\mathrm{tan}}^{-1}(\frac{2x}{1-{x}^{2}})+{\mathrm{cot}}^{-1}(\frac{1-{x}^{2}}{2x})=\frac{\pi }{3}$tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​ where $x\in (-1,1)$x∈(−1,1)

Case $1$1: When $x<0$x<0, then

$${\mathrm{tan}}^{-1}(\frac{2x}{1-{x}^{2}})+{\mathrm{cot}}^{-1}(\frac{1-{x}^{2}}{2x})=\frac{\pi }{3}$$tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​

$$\Rightarrow 2{\mathrm{tan}}^{-1}x+\pi +2{\mathrm{tan}}^{-1}x=\frac{\pi }{3}$$⇒2tan−1x+π+2tan−1x=3π​

$$\Rightarrow 4{\mathrm{tan}}^{-1}x=-\pi +\frac{\pi }{3}$$⇒4tan−1x=−π+3π​

$$\Rightarrow 4{\mathrm{tan}}^{-1}x=-\frac{2\pi }{3}$$⇒4tan−1x=−32π​

$$\Rightarrow {\mathrm{tan}}^{-1}x=-\frac{\pi }{6}$$⇒tan−1x=−6π​

$$\Rightarrow x=-\frac{1}{\sqrt{3}}$$⇒x=−3​1​

Case $2$2: When $x\geq 0$x≥0,

$${\mathrm{tan}}^{-1}(\frac{2x}{1-{x}^{2}})+{\mathrm{cot}}^{-1}(\frac{1-{x}^{2}}{2x})=\frac{\pi }{3}$$tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​

$$\Rightarrow 2{\mathrm{tan}}^{-1}x+2{\mathrm{tan}}^{-1}x=\frac{\pi }{3}$$⇒2tan−1x+2tan−1x=3π​

$$\Rightarrow {\mathrm{tan}}^{-1}x=\frac{\pi }{12}$$⇒tan−1x=12π​

$$\Rightarrow x=2-\sqrt{3}$$⇒x=2−3​

So, sum of solution will be

$$\frac{-1}{\sqrt{3}}+2-\sqrt{3}=2-\frac{4}{\sqrt{3}}$$3​−1​+2−3​=2−3​4​

Now, on comparing with $\alpha -\frac{4}{\sqrt{3}}$α−3​4​, we get

$$\alpha =2$$α=2