Question

If ${\tan x\tan y=a}$ and ${x+y=\dfrac{\pi }{6}}$, then and satisfies 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

Hint: By using the formula as is used in the formula $\tan (x+y)=\dfrac{\tan x + \tan y}{1-\tan x\tan y}$ also remember the value of $\tan \dfrac{\pi }{6}$ as $\dfrac{1}{\sqrt{3}}$.

Complete step-by-step solution -
Here put ${x+y=\dfrac{\pi }{6}}$
As the above-stated formula is used by further simplifying quadratic is formed.
\[\tan (x+y)=\dfrac{\tan x +\tan y}{(1-\tan x.\tan y)}\]
Since ${x+y=\dfrac{\pi }{6}}$
$\tan \dfrac{\pi }{6}=\dfrac{\tan x + \tan y}{1-a} \\ $
$(1-a)\dfrac{1}{\sqrt{3}}=\tan x+\tan y \\ $
Now, are the roots of the given equation.
Let $\tan x=\alpha ,\tan y=\beta $
${{x}^{2}}-(\alpha +\beta )x+\alpha \beta =0 \\ $
${{x}^{2}}-(1-a)\dfrac{1}{\sqrt{3}}x+a=0 \\ $
$ \sqrt{3}{{x}^{2}}-(1-a)x+a\sqrt{3}=0 \\ $
The correct answer is B.
Hence the ${\sqrt{3}{{x}^{2}}-1(1-a)x+a\sqrt{3}=0}$ is the satisfied equation.
Sum of roots \[=\left( \dfrac{\text{coefficient of x}}{\text{coefficient of}\; {{{x}^{2}}}} \right)\]
Product of roots = \[=\left( \dfrac{\text{constant}}{\text{coefficient of}\; {{{x}^{2}}}} \right)\].

Note: Tangent of sum of angle is used when we have a product of two tangents to simplify it. Also sum of angles is given further simplification and substitution is used. Here we need to keep in mind the relation between the quadratic equation and the roots.