Question

tanx +tan2x+root3tanx.tan2x =root3

Answer 1

Hope it's helpful to uh..

Answer 2

Given:

$tanx+tan2x+\sqrt{3}tanx\cdot tan2x=\sqrt{3}$

To find:

Value of x

Solution:

$tanx+tan2x+\sqrt{3}tanx\cdot tan2x=\sqrt{3}$

$tanx+tan2x=\sqrt{3}-\sqrt{3}tanx\cdot tan2x$

$tan+tan2x=\sqrt{3}(1-tanx\cdot tan2x)$

$\frac{tan+tan2x}{1-tanx\cdot tan2x}=\sqrt{3}$

We know that

$tan(x+y)=\frac{tanx+tany}{1-tanx\cdot tany}$

Using the formula

$tan(x+2x)=\sqrt{3}$

$tan(3x)=\sqrt{3}$

$tan(3x)=tan\frac{\pi}{3}$

We know that

$tanx=tany$

Then , $x=n\pi+y$

Using the formula

$3x=n\pi+\frac{\pi}{3}$

$x=\frac{n\pi}{3}+\frac{\pi}{9}$