Statement (A): $\cos \theta \cos (300°-\theta )\cos (300°+\theta )=\frac{1}{3}\cos 3\theta$

Statement (B): $\tan \theta \tan (60°-\theta )\tan (60°+\theta )=\frac{1}{4}\tan 3\theta$

Statement -1

$\cos \theta \cos (300°-\theta )\cos (300°+\theta )=\frac{1}{3}\cos 3\theta$

$=\cos \theta ({\cos }^{2}300°-{\sin }^2\theta )$

$=\cos \theta ({\cos }^2(270°+30°)-{\sin }^2\theta$

$=\cos \theta ({\sin }^{2}30°-{\sin }^2\theta )$

$=\cos \theta (\frac{1}{4}-{\sin }^2\theta )$

$=\cos \theta \frac{(1-4{\sin }^2\theta )}{4}$

$\therefore {\sin }^2\theta =1-{\cos }^2\theta$

$=\cos \theta \frac{[1-4+4{\cos }^2\theta ]}{4}$

$\Rightarrow \frac{1}{4}.(4{\cos }^3\theta -3\cos \theta )$

$=\frac{1}{4}\left(\cos 3\theta \right)\ne \frac{1}{3}\left(\cos 3\theta \right)$

Statement-2

$\tan \theta \left[\tan (60°-\theta )\tan (60°+\theta )\right]$

$\tan \theta \left[\frac{{\tan }^{2}60°-{\tan }^2\theta }{1-{\tan }^{2}60°{\tan }^2\theta }\right]$

$\therefore \tan (A+B)=\frac{{\tan }^{2}A-{\tan }^{2}B}{1-{\tan }^{2}A\tan B}$

$\tan \theta \left[\frac{3-{\tan }^2\theta }{1-3{\tan }^2\theta }\right]$

$=\frac{3\tan \theta -{\tan }^3\theta }{1-{\tan }^2\theta }$

$\tan 3\theta \ne \frac{1}{4}\tan 3\theta$$\left(F\right)$

Both statements are false