inverse trigonometry

\hspace{-16}$Here We Can Write The Given Series as\\\\\\ $\bf{\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{7}\right)+\tan^{-1}\left(\frac{1}{13}\right)+.........}$\\\\\\ $\bf{\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\tan^{-1}\left(\frac{1}{r^2+r+1}\right)}$\\\\\\ $\bf{\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\tan^{-1}\left(\frac{(r+1)-r}{1+r.(r+1)}\right)=\lim_{n\rightarrow \infty}\sum_{r=1}^{n}\left\{\tan^{-1}(r+1)-\tan^{-1}(r)\right\}}$\\\\\\ Now Using Telescopic Sum,We Get\\\\\\ $\bf{\lim_{n\rightarrow \infty}\left\{\tan^{-1}(n+1)-\tan^{-1}(1)\right\}}$\\\\\\ $\bf{\lim_{n\rightarrow \infty}\tan^{-1}\left\{\frac{(n+1)-1}{1+1.(n+1)}\right\}}$\\\\\\ $\bf{\lim_{n\rightarrow \infty}\tan^{-1}\left(\frac{n}{n+2}\right)}$\\\\\\ So $\bf{\tan^{-1}(1)=\tan^{-1}\tan\left\{\left(\frac{\pi}{4}\right)\right\}=\frac{\pi}{4}}$

\hspace{-16}$We Know That .......\\\\\\ $\bf{\sin^{-1}(\sin (x))=\begin{Bmatrix} \bf{x}\;\;\;\;, -\frac{\pi}{2}\leq x \leq \frac{\pi}{2} \\\\ \bf{\pi-x}\;\;\;\;, \frac{\pi}{2}\leq x \leq \frac{3\pi}{2} \\\\ \bf{x-2\pi}\;\;\;\;, \frac{3\pi}{2}\leq x \leq \frac{5\pi}{2}\\\\ \bf{3\pi-x}\;\;\;\;, \frac{5\pi}{2}\leq x \leq \frac{7\pi}{2}\\\\ \bf{x-4\pi}\;\;\;\;, \frac{7\pi}{2}\leq x \leq \frac{9\pi}{2} \end{Bmatrix}}$\\\\\\ So $\bf{\sin^{-1}\left(\sin(12)\right)=12-4\pi}$\\\\\\ Similarly.......\\\\\\ $\bf{\cos^{-1}\left(\cos (x)\right)=\begin{Bmatrix} \bf{x}\;\;\;,0\leq x \leq \pi \\\\ \bf{2\pi-x}\;\;\;,\pi\leq x \leq 2\pi\\\\ \bf{x-2\pi}\;\;\;,2\pi\leq x \leq 3\pi\\\\ \bf{4\pi-x}\;\;\;,3\leq x \leq 4\pi \\\\ \end{Bmatrix}}$\\\\\\ So $\bf{\cos^{-1}\left(\cos (12)\right)=4\pi-12}$\\\\\\ So $\bf{\sin^{-1}\left(\sin (12)\right)+\cos^{-1}\left(\cos (12)\right)=12-4\pi+4\pi-12=0}$

awesome... why couldn't i think of that...!!

well for the second one i meant 4pi not 2pi ... actually...

well thanks a tonne sir...!!