Question

value of tan 2 π/3
slove by using this formula, tan(A+B) = tana +tanb/1-tana . tanb ​

Answer 1

Appropriate Question :-

$\rm \: Find \: the \: value \: of \: tan\dfrac{2\pi}{3}, \: using \: the \: formula \:$

$\rm \: tan(a + b) = \dfrac{tana + tanb}{1 - tana \: tanb}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: tan(a + b) = \dfrac{tana + tanb}{1 - tana \: tanb}$

Now, Substitute

$\rm \: a \: = \: \dfrac{\pi}{3}$

$\rm \: b \: = \: \dfrac{\pi}{3}$

we get

$\rm \: tan\bigg(\dfrac{\pi}{3} + \dfrac{\pi}{3} \bigg) = \dfrac{tan\dfrac{\pi}{3} + tan\dfrac{\pi}{3}}{1 - tan\dfrac{\pi}{3} \times tan\dfrac{\pi}{3}}$

$\rm \: tan\dfrac{2\pi}{3} = \dfrac{ \sqrt{3} + \sqrt{3} }{1 - \sqrt{3} \times \sqrt{3} }$

$\rm \: tan\dfrac{2\pi}{3} = \dfrac{ 2\sqrt{3}}{1 - 3}$

$\rm \: tan\dfrac{2\pi}{3} = \dfrac{ 2\sqrt{3}}{ -2 }$

$\rm\implies \boxed{\sf{  \:\rm \: \:tan\dfrac{2\pi}{3} = - \sqrt{3} \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\boxed{\sf{  \:\rm \: sin(x + y) = sinx \: cosy \: + \: siny \: cosx \: \: }}$

$\boxed{\sf{  \:\rm \: sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: \: }}$

$\boxed{\sf{  \:\rm \: cos(x + y) = cosx \: cosy \: - \: sinx \: siny \: \: }}$

$\boxed{\sf{  \:\rm \: cos(x - y) = cosx \: cosy \: + \: sinx \: siny \: \: }}$

$\boxed{\sf\rm \: {  \:tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany} \: \: }}$