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 : $sin 18° =\frac{\sqrt{5}-1}{4}$
Answer & explanation
Correct answer: option 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.