Question

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

• A. $0$
• B. $\pi /4$
• C. $1$
• D. $\pi /2$

Answer 1

The correct option is B

$\pi /4$

Find the value of $\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\beta }$:

Given that $\alpha$ and $\beta$are the roots of the equations$6x^2-5x+1=0$,

$\Rightarrow (\alpha +\beta )=\frac{-(-5)}{6}=\frac{5}{6},\alpha \beta =\frac{1}{6}$

$$\begin{gathered} \mathbf{} \\ \Rightarrow {\tan }^{-1}\alpha +{\tan }^{-1}\beta ={\tan }^{-1}\left[\frac{\left({\displaystyle \frac{5}{6}}\right)}{\left(1-{\displaystyle \frac{1}{6}}\right)}\right]\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathit{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{}\mathit{\beta }\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left[}\frac{\mathbf{(}\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\beta }\mathbf{)}\mathbf{}}{\mathbf{(}\mathbf{1}\mathbf{}\mathbf{–}\mathbf{}\mathbf{\alpha \beta }\mathbf{)}}\mathbf{\right]}\mathbf{} \\ \Rightarrow {\tan }^{-1}\alpha +{\tan }^{-1}\beta ={\tan }^{-1}\left(1\right) \\ \Rightarrow {\tan }^{-1}\alpha +{\tan }^{-1}\beta =\pi /4 \end{gathered}$$

Hence, the correct option is (B).