Solve for $x$ if $\tan^{-1}{\frac{\sqrt{1+x^2}-1}{x}}=\frac{\pi}{45}$

Question

It given that $$\tan^{-1}{\frac{\sqrt{1+x^2}-1}{x}}=\frac{\pi}{45}$$ Solve for $x$

So I am pretty new to inverse trigonometric functions so I don't really know what to do over here. Please help

Answer 1

Apply the function $\tan$ to both sides

$$ \frac{\sqrt{1 + x^2} - 1}{x} = \tan(\pi/45) $$

Rearranging

$$ 1 + x^2 = [1 + x\tan(\pi/45)]^2 $$

Expand this last term, and solve the quadratic equation. The result is

$$ x = \frac{2\tan(\pi/45)}{1 - \tan^2(\pi/45)} $$

If you apply a double angle formula for the tangent, this is equivalent to

$$ x = \tan(2\pi/45) $$

Answer 2

Let $x=\tan y$ where $-\dfrac\pi2<y<\dfrac\pi2$

$\dfrac{\sqrt{1+x^2}-1}x=\dfrac{+\sec y-1}{\tan y}$ as $\sec y>0$

$\implies\dfrac{\sqrt{1+x^2}-1}x=\tan\dfrac y2$

$\implies\dfrac y2=\dfrac\pi{45}$

$x=?$