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The value of $\sum\limits^{3}_{n=1} tan^{-1}\frac{1}{n}$, is |
0 $\pi $ $\frac{\pi}{2}$ none of these |
$\frac{\pi}{2}$ |
$\sum\limits^{3}_{n=1} tan^{-1}\frac{1}{n}= tan^{-1}+tan^{-1}\frac{1}{2}+tan^{-1}\frac{1}{3}$ $\sum\limits^{3}_{n=1}tan^{-1}\frac{1}{n}=\frac{\pi}{4}+tan^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}×\frac{1}{3}}\right)$ $\sum\limits^{3}_{n=1}tan^{-1}\frac{1}{n}=\frac{\pi}{4}+tan^{-1}1=\frac{\pi}{4}+\frac{\pi}{4}=\frac{\pi}{2}$ |
