Question

y=tan-1(tanθ−tanπ41+tanθtanπ4)

Answer 1

$y = \begin{cases} \tan^{-1} X - \frac{\pi}{4} & \text{if } X > -1 \\ \tan^{-1} X + \frac{3\pi}{4} & \text{if } X \le -1 \end{cases}$

Answer 2

The given expression for $y$ is simplified using the tangent subtraction formula: $y = \tan^{-1}\left(\frac{\tan \theta - \tan \frac{\pi}{4}}{1 + \tan \theta \tan \frac{\pi}{4}}\right) = \tan^{-1}\left(\tan\left(\theta - \frac{\pi}{4}\right)\right)$ Assuming $X = \tan \theta$, we have $\theta = \tan^{-1} X$, where $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Then $y = \tan^{-1}\left(\tan\left(\tan^{-1} X - \frac{\pi}{4}\right)\right)$. Let $Z = \tan^{-1} X - \frac{\pi}{4}$. The range of $Z$ is $(-\frac{3\pi}{4}, \frac{\pi}{4})$. Using $\tan^{-1}(\tan Z) = Z - n\pi$, where $Z - n\pi \in (-\frac{\pi}{2}, \frac{\pi}{2})$: If $X > -1$, then $\tan^{-1} X > -\frac{\pi}{4}$, so $Z \in (-\frac{\pi}{4}, \frac{\pi}{4})$. Here $n=0$, so $y = Z = \tan^{-1} X - \frac{\pi}{4}$. If $X \le -1$, then $\tan^{-1} X \le -\frac{\pi}{4}$, so $Z \in (-\frac{3\pi}{4}, -\frac{\pi}{4}]$. Here $n=-1$, so $y = Z + \pi = \tan^{-1} X - \frac{\pi}{4} + \pi = \tan^{-1} X + \frac{3\pi}{4}$.