Question

If $\alpha$ and $\beta$ are the roots of the equation $6x^2-5x+1=0$, then the value of ${\tan }^{-1}\left(\alpha \right)+{\tan }^{-1}\left(\beta \right)$ is

• A. $0$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $1$
• D. $\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Step 1: Find the sum and product of the roots of the given quadratic equation.

We have given a quadratic equation $6x^2-5x+1=0$, whose roots are $\alpha$ and $\beta$.

$\begin{array}{rcl}\alpha +\beta & =& \frac{-b}{a}\\ \alpha +\beta & =& \frac{-(-5)}{6}\\ \alpha +\beta & =& \frac{5}{6}\\ \alpha \beta & =& \frac{c}{a}\\ \alpha \beta & =& \frac{1}{6}\end{array}$

Step 2: Find the value of $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\alpha \right)\mathbf{+}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\beta \right)$

so,

$\begin{array}{rcl}{\tan }^{-1}\left(\alpha \right)+{\tan }^{-1}\left(\beta \right)& =& ta{n}^{-1}\left(\frac{{\displaystyle \frac{5}{6}}}{1-{\displaystyle \frac{1}{6}}}\right)\mathbf{\{}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\alpha \right)\mathbf{+}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\beta \right)\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left(\frac{{\displaystyle \alpha +\beta }}{1-{\displaystyle \mathrm{\alpha \beta }}}\right)\}\\ & =& ta{n}^{-1}\left(\frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{5}{6}}}\right)\\ & =& ta{n}^{-1}\left(1\right)\\ & =& \frac{\mathrm{\pi }}{4}\end{array}$

${\tan }^{-1}\left(\alpha \right)+{\tan }^{-1}\left(\beta \right)=\frac{\mathrm{\pi }}{4}$

Hence, the correct option is (B).