phool chandra
Solution:
We know:-
$ tan^{-1}\sqrt{3} = \frac{\pi}{3}$
$cot^{-1}\sqrt{3} = \frac{\pi}{6}$
$ tan^{-1}\sqrt{3} - cot^{-1}\sqrt{3} $ = $ \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6} $
So, $ tan^{-1}\sqrt{3} - cot^{-1}\sqrt{3} = \frac{\pi}{6}$
When $a \ne 0$, there are two solutions to \(ax^2 + bx + c = 0\) and they are
$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$
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Find the value of $ tan^{-1}\sqrt{3} - cot^{-1}\sqrt{3} $.
Sept. 24, 2023
We know:-
$ tan^{-1}\sqrt{3} = \frac{\pi}{3}$
$cot^{-1}\sqrt{3} = \frac{\pi}{6}$
$ tan^{-1}\sqrt{3} - cot^{-1}\sqrt{3} $ = $ \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6} $
So, $ tan^{-1}\sqrt{3} - cot^{-1}\sqrt{3} = \frac{\pi}{6}$