Question

If $x>1$ then $2{\tan }^{-1}x$ cannot be equal to

Answer 1

Compute the required value:

Let $a={\tan }^{-1}x$

then $x=\tan a$

therefore,

$\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{2}\mathit{a}\mathbf{=}\mathbf{2}\mathit{c}\mathit{o}{\mathit{s}}^{\mathbf{2}}\mathit{a}\mathbf{-}\mathbf{1}$

$$\begin{gathered} \Rightarrow \cos 2\mathrm{a}=2{\left(\frac{1}{\sqrt{1+{\mathrm{x}}^2}}\right)}^2-1 \\ \Rightarrow \cos 2\mathrm{a}=\left(\frac{2}{1+{\mathrm{x}}^2}\right)-1 \\ \Rightarrow \cos 2\mathrm{a}=\left(\frac{2-(1+{\mathrm{x}}^2)}{1+{\mathrm{x}}^2}\right) \\ \Rightarrow \cos 2\mathrm{a}=\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2} \\ \Rightarrow 2\mathrm{a}={\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right).....\left(1\right) \end{gathered}$$

We know that

$a={\tan }^{-1}x$

it can be written as

$2a=2{\tan }^{-1}x.....\left(2\right)$

From equation $\left(1\right)and\left(2\right)$, we can write that

$2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)$

Hence, if $x>1$ then $2{\tan }^{-1}x\ne 2a={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)$.