Question

The value of $3\,tan^610^{\circ} - 27\,tan^410^{\circ} + 33 \,tan^210^{\circ}$ equals

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

Answer 1

Answer: (C)

$tan\,3A=\frac{3\,tan\,A-tan^{3}\,A}{1-3\,tan^{2}\,A}\quad\ldots\left(i\right)$
Putting $A = 10^{\circ}$ in $\left(i\right)$, we get
$\therefore \frac{1}{\sqrt{3}}\left(1-3\,tan^{2}\,10^{\circ}\right)=tan 10^{\circ}\left(3 - tan^{2}10^{\circ}\right)\quad\ldots\left(ii\right)$
Squaring $\left(ii\right)$ both sides, we get
$\left(1 -3\, tan^{2}10^{\circ}\right)^2 = 3\, tan^{2}10^{\circ} \left(3 - tan^{2}10^{\circ}\right)^{2}$
$\Rightarrow 3\, tan^{2}10^{\circ} \left(9 + tan^{4}10^{\circ} - 6tan^{2}10^{\circ}\right)$
$= 1 + 9 \,tan^410^{\circ} - 6tan^210^{\circ}$
$\Rightarrow 3 \,tan^{6}10^{\circ} - 27\, tan^{4}10^{\circ} + 33 \,tan^{2}10^{\circ}= 1$