Question

If $tanx\cdot tany=a$ 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. $\sqrt{3}{x}^2-(1-a)x+a\sqrt{3}=0$
• C. $x^2+\sqrt{3}(1+a)x-a=0$
• D. $\sqrt{3}{x}^2+(1+a)x-a\sqrt{3}=0$

Answer 1

Answer: (B)

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

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

$\tan \frac{\Pi }{6}=\frac{\tan x+\tan y}{1-a}$

$\frac{1}{\sqrt{3}}=\frac{\tan x+\tan y}{1-a}$

$\tan x+\tan y=\frac{1-a}{\sqrt{3}}$

Let $\tan x,\tan y$ be the root of quadratic equation

$x^2-(\tan x+\tan y)x+\left(\tan x\tan y\right)=0$

$x^2-\frac{1-a}{\sqrt{3}}x+a=0$

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

Ans will be option B