Question

If $ \tan x.\tan y=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)

$\because \tan x \cdot \tan y=a$
and $\tan (x+y)=\tan \left(\frac{\pi}{6}\right)$
$\Rightarrow \frac{\tan x+\tan y}{1-\tan x \cdot \tan y}=\frac{1}{\sqrt{3}}$
$\Rightarrow \tan x+\tan y=\frac{1}{\sqrt{3}}(1-a)$
Equation whose roots are tan $x$ and tan $y$ is
$x^{2}-\frac{(1-a)}{\sqrt{3}} \cdot x+a=0$
$\Rightarrow \sqrt{3} x^{2}-(1-a) x+a \sqrt{3}=0$