Question

If $\alpha $ and $\beta $ are the roots of the equations $6{{x}^{2}}-5x+1=0$, then the value of ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $ is
A . 0
B. $\dfrac{\pi }{4}$
C. 1
D. $\dfrac{\pi }{2}$

Answer 1

Hint: In this question, we are given a quadratic equation with its roots and we have to find the value of ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $. First, we compare the given equation with the standard form of a quadratic equation and then find the sum of roots and the product of roots of the given quadratic equation. Then we put the values in the formula of ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $ and find the value and choose the correct option.

Formula used:
Sum of roots = $-\dfrac{b}{a}$
And the product of roots = $\dfrac{c}{a}$
And ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $= ${{\tan }^{-1}}\left[ \dfrac{\left( \alpha +\beta \right)}{\left( 1-\alpha \beta \right)} \right]$

Complete step by step Solution:
Given quadratic equation is $6{{x}^{2}}-5x+1=0$…………………….. (1)
Compare it with standard form of quadratic equation $a{{x}^{2}}+bx+c=0$, we get
$a=6,b=-5,c=1$
Consider $\alpha $and $\beta $ are the roots of the equation (1)
then the sum of roots ($\alpha +\beta $) = $-\dfrac{b}{a}$= $-\dfrac{(-5)}{6}$= $\dfrac{5}{6}$
and the product of roots ($\alpha \beta $) = $\dfrac{c}{a}$= $\dfrac{1}{6}$
As we have to find the value of ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $
We know ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $= ${{\tan }^{-1}}\left[ \dfrac{\left( \alpha +\beta \right)}{\left( 1-\alpha \beta \right)} \right]$
Now we put the value of $\alpha +\beta $and $\alpha \beta $in the above equation, we get
${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $= ${{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{5}{6} \right)}{1-\dfrac{1}{6}} \right]$
Solving further, we get
${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $= ${{\tan }^{-1}}\left[ \dfrac{\left( \dfrac{5}{6} \right)}{\dfrac{5}{6}} \right]$
That is ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $= ${{\tan }^{-1}}\left( 1 \right)$
We know $\tan \dfrac{\pi }{4}=1$
Then ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $= ${{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)$
Hence ${{\tan }^{-1}}\alpha +{{\tan }^{-1}}\beta $ = $\dfrac{\pi }{4}$

Therefore, the correct option is (B).

Note: In these types of questions, we can find the sum and the product of roots by the formula
 if x and y are the roots of any quadratic equation the value of xy will be equal to $\dfrac{ constant\, term}{coefficient\, of\, x^2}$ and the sum of the roots that is x + y is equal to $\dfrac{ -coefficient\, of \,x}{coefficient\, of\, x^2}$ and by solving it we get the desired answer.