Question

If $co{t}^{-1}\left(\alpha \right)+co{t}^{-1}\left(\beta \right)=co{t}^{-1}\left(x\right)$, then $x$

• A. $\alpha +\beta$
• B. $\alpha –\beta$
• C. $\frac{(1+\alpha \beta )}{(\alpha +\beta )}$
• D. $\frac{(\alpha \beta –1)}{(\alpha +\beta )}$

Answer 1

The correct option is D

$\frac{(\alpha \beta –1)}{(\alpha +\beta )}$

Find the value of $x$:

Given, $co{t}^{-1}\left(\alpha \right)+co{t}^{-1}\left(\beta \right)=co{t}^{-1}\left(x\right)$

$\Rightarrow$${\tan }^{-1}\left(\frac{1}{\alpha }\right)+{\tan }^{-1}\left(\frac{1}{\beta }\right)={\tan }^{-1}\left(\frac{1}{x}\right)$

$\Rightarrow$ $ta{n}^{-1}\left[\frac{\left({\displaystyle \frac{1}{\alpha }}\right)+\left({\displaystyle \frac{1}{\beta }}\right)}{1–\left({\displaystyle \frac{1}{\alpha }}\right)\left({\displaystyle \frac{1}{\beta }}\right)}\right]=ta{n}^{-1}\left(\frac{1}{x}\right)$ $\left[\mathbf{\because }\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\frac{a+b}{1–ab}\right)\right]$

$\Rightarrow$ ${\tan }^{-1}\left(\frac{\beta +\alpha }{\alpha \beta –1}\right)={\tan }^{-1}\left(\frac{1}{x}\right)$

$\Rightarrow$ $\left(\frac{\alpha +\beta }{\alpha \beta –1}\right)=\left(\frac{1}{x}\right)$

$\mathbf{\therefore }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{(}\mathbf{\alpha }\mathbf{\beta }\mathbf{}\mathbf{–}\mathbf{}\mathbf{1}\mathbf{)}}{\mathbf{(}\mathbf{\alpha }\mathbf{}\mathbf{+}\mathbf{}\mathbf{\beta }\mathbf{)}}$

Hence, Option 'D is Correct.