Inverse trigonometry. Write this function in the simplest form.

Question

It is a question from NCERT class 12 maths textbook. From chapter 2.exercise 2.2 question no. 5(https://i.stack.imgur.com/takqu.png)

$$\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}$$

Answer 1

Hint :

Let $$x=\tan \theta \,\, ; \,\, \theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$$ and see what happens.

Answer 2

This inverse trigonometric function can be solved by substituting $x=\tan \theta$

Henceforth, we obtain, $$ \tan^{-1} \frac{\sec\theta -1}{\tan\theta}$$ $$=\tan^{-1} \frac{1-\cos\theta}{\sin\theta}$$

At this juncture, you may apply the half angle formulae, i.e, $\sin\theta= 2\sin \frac{\theta}{2}\cos\frac{\theta}{2}$ and $\cos\theta=1-2\sin^2\frac{\theta}{2}$,simplify thereafter to reach the final result.

Note: Taking the negative sign for $\sec\theta$ gives a second solution

In that case, we obtain$$ \tan^{-1} \frac{-(\sec\theta +1)}{\tan\theta}$$ $$=\tan^{-1} \frac{-(1+\cos\theta)}{\sin\theta}$$

Which can again be simplified with the half angle formulae to obtain the second solution.

Answer 3

$$ \frac{\sqrt{1+x^2}-1}{x}$$ is $1-1$ function $$y=\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}\to \color{red} {***frac{-\pi}{2}<y<\frac{+\pi}{2}}\\ \tan(y)=tan(\tan^{-1} \frac{\sqrt{1+x^2}-1}{x})\\ \tan y =\frac{\sqrt{1+x^2}-1}{x}\\ \sqrt{1+x^2}=x\tan y +1\\$$to the power of two $$1+x^2=x^2\tan^2 y +2x\tan y +1 \\ \to x\neq 0 \\x=\dfrac{-2\tan y }{\tan^2 y -1} \to \\ f^{-1}(x)=\dfrac{-2\tan x }{\tan^2 x -1}\\ \to \color{red} {***}\frac{-\pi}{2}<x<\frac{+\pi}{2}$$