Statement-1: if $ x ∈ [-1 / \sqrt{3}, 1/ \sqrt{3}],$ then

$cot^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) = cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) ⇒ x = \frac{\sqrt{25-10\sqrt{5}}}{5}$

Statement-2 : $sin 18° =\frac{\sqrt{5}-1}{4}$

Statement 1 is True, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

Clearly, statement-2 is true.

Now,  $cot^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) = cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$⇒ \frac{\pi}{2} - tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right) = cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)$

$⇒\frac{\pi}{2} - 3tan^{-1} x = 2 tan^{-1}x$          $[∵-\frac{1}{\sqrt{3}}≤ x ≤ \frac{1}{\sqrt{3}}]$

$⇒ tan^{-1}x= \frac{\pi}{10}$

$⇒ x = tan \frac{\pi}{10}=\frac{sin18°}{cos18°}=\frac{\sqrt{5}-1}{\sqrt{10+2\sqrt{5}}}$

$⇒ x = \frac{(\sqrt{5}-1)\sqrt{10-2\sqrt{5}}}{\sqrt{100-20}}⇒ x = \frac{\sqrt{(10-2\sqrt{5})(6-2\sqrt{5})}}{4\sqrt{5}}$

$⇒ x = \frac{\sqrt{80-32\sqrt{5}}}{4\sqrt{5}}-\frac{\sqrt{5-2\sqrt{5}}}{\sqrt{5}}-\frac{\sqrt{25-10\sqrt{5}}}{5}$

So, statement-1 is true.

Also, statement-2 is a correct explanation for statement-1.