Question

If $tanxtany=a$ and $x+y=$ $\frac{\pi }{6}$, then $\tan x$ and $\tan y$ satisfy the equation

• A. $x^2-\sqrt{3}(1-a)x+a=0$
• B. $\sqrt{3}{x}^2-1(1-a)x+a\sqrt{3}=0$
• C. $a^2\sqrt{3}(1+a)x-a=0$
• D. $\sqrt{3}{x}^2+(1+a)x-a\sqrt{3}=0$

Answer 1

The correct option is B $\sqrt{3}{x}^2-1(1-a)x+a\sqrt{3}=0$

$x+y=\frac{\pi }{6}$

$\Rightarrow tan(x+y)=tan\left(\frac{\pi }{6}\right)$

$\Rightarrow \frac{tanx+tany}{1-tanx.tany}=\frac{1}{\sqrt{3}}$

$\Rightarrow tanx+tany=\frac{1}{\sqrt{3}}(1-tanx.tany)=\frac{1}{\sqrt{3}}(1-a)$

So equation with roots $tanx,tany$ is

$x^2-(tanx+tany)x+(tanx.tany)=0$

$\Rightarrow \sqrt{3}{x}^2-(1-a)+a\sqrt{3}=0$