Mathematics - Limits Question with Solution | TestHub

${\cot }^{-1}\left(r^2+\frac{3}{4}\right)={\tan }^{-1}\left(\frac{1}{r^2+\frac{3}{4}}\right)$

$$\begin{array}{rl}& ={\tan }^{-1}\left(\frac{1}{1+\left(r^2-\frac{1}{4}\right)}\right)\\ & ={\tan }^{-1}\left(\frac{1}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right)\\ & ={\tan }^{-1}\left(\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r^2+\frac{1}{4}\right)}\right)\\ & ={\tan }^{-1}\left(r+\frac{1}{2}\right)-{\tan }^{-1}\left(r-\frac{1}{2}\right)\end{array}$$