Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Let $t=\tan x$. We use the identity $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$.$\tan (x+\pi /3)=\frac{\tan x+\tan (\pi /3)}{1-\tan x\tan (\pi /3)}=\frac{t+\sqrt{3}}{1-\sqrt{3}t}$.$\tan (x+2\pi /3)=\frac{\tan x+\tan (2\pi /3)}{1-\tan x\tan (2\pi /3)}=\frac{t-\sqrt{3}}{1+\sqrt{3}t}$. For option (A): The sum $S=\tan x+\tan (x+\pi /3)+\tan (x+2\pi /3)=t+\frac{t+\sqrt{3}}{1-\sqrt{3}t}+\frac{t-\sqrt{3}}{1+\sqrt{3}t}$.$$S=t+\frac{(t+\sqrt{3})(1+\sqrt{3}t)+(t-\sqrt{3})(1-\sqrt{3}t)}{(1-\sqrt{3}t)(1+\sqrt{3}t)}=t+\frac{(t+\sqrt{3}{t}^2+\sqrt{3}+3t)+(t-\sqrt{3}{t}^2-\sqrt{3}+3t)}{1-3t^2}$$ $$S=t+\frac{8t}{1-3t^2}=\frac{t(1-3t^2)+8t}{1-3t^2}=\frac{9t-3t^3}{1-3t^2}=3\left(\frac{3t-t^3}{1-3t^2}\right)$$ We know that $\tan 3x=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}=\frac{3t-t^3}{1-3t^2}$. Therefore, $S=3\tan 3x$. (A) is correct. For option (B): The product $P=\tan x\tan (x+\pi /3)\tan (x+2\pi /3)=t\cdot \left(\frac{t+\sqrt{3}}{1-\sqrt{3}t}\right)\cdot \left(\frac{t-\sqrt{3}}{1+\sqrt{3}t}\right)$.$$P=t\cdot \frac{t^2-3}{1-3t^2}=-t\cdot \frac{3-t^2}{1-3t^2}=-\left(\frac{3t-t^3}{1-3t^2}\right)=-\tan 3x$$ So, $\tan x\tan (x+\pi /3)\tan (x+2\pi /3)+\tan 3x=0$. (B) is correct. For option (C): From (A), if $\tan 3x=0$, then $\tan x+\tan (x+\pi /3)+\tan (x+2\pi /3)=3(0)=0$. (C) is correct. For option (D): From (B), if $\tan x\tan (x+\pi /3)\tan (x+2\pi /3)=1$, then $1+\tan 3x=0$, which implies $\tan 3x=-1$. (D) is correct.