Question

Solve tan(120+A)+tan(120-A)=¿

Answer 1

Answer:

$\tan \left(120^{\circ \:}+A\right)+\tan \left(120^{\circ \:}-A\right)=\tan \left(\dfrac{360^{\circ \:}+3A}{3}\right)+\tan \left(\dfrac{360^{\circ \:}-3A}{3}\right)$

Step-by-step explanation:

$\tan \left(120^{\circ \:}+A\right)+\tan \left(120^{\circ \:}-A\right)$

$\mathrm{Join}\:120^{\circ \:}+A$

$120^{\circ \:}+A$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:A=\dfrac{A3}{3}$

$=120^{\circ \:}+\dfrac{A\cdot \:3}{3}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{360^{\circ \:}+A\cdot \:3}{3}$

$=\tan \left(\dfrac{3A+360^{\circ \:}}{3}\right)+\tan \left(120^{\circ \:}-A\right)$

$\mathrm{Join}\:120^{\circ \:}-A$

$120^{\circ \:}-A$

$\mathrm{Convert\:element\:to\:fraction}:\quad \:A=\dfrac{A3}{3}$

$=120^{\circ \:}-\dfrac{A\cdot \:3}{3}$

$\mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}:\quad \dfrac{a}{c}\pm \dfrac{b}{c}=\dfrac{a\pm \:b}{c}$

$=\dfrac{360^{\circ \:}-A\cdot \:3}{3}$

$=\tan \left(\dfrac{3A+360^{\circ \:}}{3}\right)+\tan \left(\dfrac{360^{\circ \:}-3A}{3}\right)$