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prove that $(\tan ^{-1}(1)+\tan ^{-1}(2)+\tan ^{-1}(3)=\pi$
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Solution:

$\tan ^{-1}(1)+\tan ^{-1}(2)+\tan ^{-1}(3)$

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

$=\pi+\tan ^{-1}(-3)+\tan ^{-1}(3)$

$=\pi-\tan ^{-1}(3)+\tan ^{-1}(3)$

$=\pi$

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