Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$\tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}$. For $x\ne \frac{k\pi }{6}$ : $\frac{\tan 3x}{\tan x}=\frac{3-{\tan }^{2}x}{1-3{\tan }^{2}x}=\frac{(\sqrt{3}-\tan x)(\sqrt{3}+\tan x)}{(1+\sqrt{3}\tan x)(1-\sqrt{3}\tan x)}=\tan \left(\frac{\pi }{3}-x\right)\tan \left(\frac{\pi }{3}+x\right)$, so (D) is correct.

At $x=\frac{\pi }{6}$ (and every $x=\frac{\pi }{6}+\frac{k\pi }{3}$) both sides are undefined, so the relation is not an identity there. Option (c) allows $x=\frac{\pi }{6}$, so (c) is false. Only (d) is correct.