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Value of $tan^{-1}2+tan^{-1}3+tan^{-1}4$ is equal to : |
$\pi +tan^{-1}\frac{3}{5}$ $\pi - tan^{-1}\frac{3}{5}$ $\pi - tan^{-1}\frac{5}{3}$ $\pi + tan^{-1}\frac{5}{3}$ |
$\pi +tan^{-1}\frac{3}{5}$ |
$\tan^{-1}2+\tan^{-1}3+\tan^{-1}4$ $=π+\tan^{-1}\left(\frac{2+3}{1-2×3}\right)+\tan^{-1}4$ (π add as 1-2.3<0) so $π-\tan^{-1}1+\tan^{-1}4$ $=π+\tan^{-1}\left(\frac{4-1}{1+4.1}\right)$ $=π+\tan^{-1}\left(\frac{3}{5}\right)$ |
