Question

In $△ABC$ ,the value of $\frac{tanA+tan(A+{120}^o)+tan(A+{240}^o)}{tan3A}=?$

• A. $3$
• B. $-3$
• C. $2$
• D. $1$

Answer 1

The correct option is A $3$

$tan(A+{120}^{\circ })=\frac{tanA+tan{120}^{\circ }}{1-tanA\times tan{120}^{\circ }}$

$tan(A+{240}^{\circ })=\frac{tanA+tan{240}^{\circ }}{1-tanA\times tan{240}^{\circ }}$

We know that

$tan{120}^{\circ }=tan({90}^{\circ }+{30}^{\circ })=-cot{30}^{\circ }=-\sqrt{3}$

$tan{240}^{\circ }=tan({180}^{\circ }+{60}^{\circ })=tan{60}^{\circ }=\sqrt{3}$

$tan3A=\frac{3tanA-ta{n}^{3}A}{1-3ta{n}^{2}A}$

So,

$tan(A+{120}^{\circ })=\frac{tanA-\sqrt{3}}{1+\sqrt{3}tanA}$

$tan(A+{240}^{\circ })=\frac{tanA+\sqrt{3}}{1-\sqrt{3}tanA}$

Therefore,

$\frac{tanA+tan(A+{120}^{\circ })+tan(A+{240}^{\circ })}{tan3A}=\frac{tanA+\frac{tanA-\sqrt{3}}{1+\sqrt{3}tanA}+\frac{tanA+\sqrt{3}}{1-\sqrt{3}tanA}}{tan3A}$

$=\frac{tanA(1-3ta{n}^{2}A)+(tanA-\sqrt{3})(1-\sqrt{3}tanA)+(tanA+\sqrt{3})(1+\sqrt{3}tanA)}{tan3A(1-3ta{n}^{2}A)}$

$=\frac{tanA-3ta{n}^{3}A+8tanA}{tan3A(1-3ta{n}^{2}A)}=\frac{9tanA-3ta{n}^{3}A}{tan3A(1-3ta{n}^{2}A)}=\frac{3tan3A}{tan3A}=3$