Practicing Success
$tan^{-1}\left(tan\sqrt{1-\theta }\right) = \sqrt{1-\theta }$ when |
$-\frac{\pi}{2} < \theta < \frac{\pi}{2}$ $\theta > \frac{4-\pi}{4}$ $\theta < \frac{4-\pi}{4}$ $ \frac{4-\pi^2}{4} < \theta ≤ 1$ |
$ \frac{4-\pi^2}{4} < \theta ≤ 1$ |
Clearly, $\sqrt{1-\theta }$ is real, if $\theta ≤ 1$ Now, $tan^{-1}\left(tan\sqrt{1-\theta }\right) = \sqrt{1-\theta }$ $ ⇒ 0 ≤\sqrt{1-\theta } < \frac{\pi}{2} $ and $ \theta ≤ 1$ $ ⇒ 0 ≤1-\theta < \frac{\pi^2}{4} $ and $ \theta ≤ 1$ $ ⇒ -1 ≤-\theta < \frac{\pi^2}{4} -1$ and $ \theta ≤ 1$ $ ⇒ 1- \frac{\pi^2}{4} < \theta ≤ 1$ and $\theta ≤ 1 ⇒\frac{4-\pi^2}{4} < \theta ≤ 1$ |