Question

The value of $3ta{n}^{6}10^{\circ }-27ta{n}^{4}10^{\circ }+33ta{n}^{2}10^{\circ }$ equals

• A. $0$
• B. $-1$
• C. $1$
• D. None of these

Answer 1

Answer: (C)

$tan3A=\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}\dots \left(i\right)$
Putting $A=10^{\circ }$ in $\left(i\right)$, we get
$\therefore \frac{1}{\sqrt{3}}\left(1-3ta{n}^{2}10^{\circ }\right)=tan10^{\circ }\left(3-ta{n}^{2}10^{\circ }\right)\dots \left(ii\right)$
Squaring $\left(ii\right)$ both sides, we get
${\left(1-3ta{n}^{2}10^{\circ }\right)}^2=3ta{n}^{2}10^{\circ }{\left(3-ta{n}^{2}10^{\circ }\right)}^2$
$\Rightarrow 3ta{n}^{2}10^{\circ }\left(9+ta{n}^{4}10^{\circ }-6ta{n}^{2}10^{\circ }\right)$
$=1+9ta{n}^{4}10^{\circ }-6ta{n}^{2}10^{\circ }$
$\Rightarrow 3ta{n}^{6}10^{\circ }-27ta{n}^{4}10^{\circ }+33ta{n}^{2}10^{\circ }=1$