Solving $\tan^{-1}x > \cot^{-1}x$

Question

I am doing problems in Inverse Trigonometric Functions. I faced some issues while solving Inverse Trigonometric Inequalities. I have mentioned the question, the solution given in my book and the way in which I attempted the problem in the image below.

As you could see, clearly the book answer does not match with mine. I tried my best to identify the mistake in my procedure, but I was unable to find any errors. Kindly tell where I have went wrong. I am sure I have gone wrong since I solved the question graphically and attained the result given in my book.

Answer 1

The problem lies in$$\arctan(x)>\pi+\arctan\left(\frac1x\right)\implies x>\tan\left(\pi+\arctan\left(\frac1x\right)\right).$$The tangent function is not a strictly increasing function, although its restriction to intervals of the form $\left(k\pi-\frac\pi2,k\pi+\frac\pi2\right)$ is.

Answer 2

Use https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

$$\dfrac\pi2>\tan^{-1}x,\cot^{-1}x>0$$

We must have $$\dfrac\pi2>\tan^{-1}x>\cot^{-1}x>0$$

$\implies x>0$