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It's said that

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

Well what about this. Lets say

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

Now let

$$2\tan^{-1}x=\tan^{-1}y\tag{2}$$

Then we can write

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

$$=-\frac{\tan^{-1}y}{2}$$

$$=-\frac{2\tan^{-1}x}{2}$$

$$=-\tan^{-1}x\neq(1)$$

What's happening? Am I wrong somewhere?

Jam
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Hijaz Aslam
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    You have said$$\frac{\left(\pi -tan^{-1}y\right)}{2}=-\frac{tan^{-1}y}{2}\ :$$a $\pi/2$ term has just disappeared. – David Aug 09 '14 at 11:33
  • $\frac{\pi }{2}-tan^{-1}x:=:\frac{\left(\pi :-2tan^{-1}x\right)}{2}$ – Hijaz Aslam Aug 09 '14 at 14:36
  • http://math.stackexchange.com/questions/587865/arctan-fracxy1%E2%88%92xy-arctan-x-arctan-y-for-this-to-be-true-xy-has and http://math.stackexchange.com/questions/610261/simplifying-an-arctan-equation – lab bhattacharjee Aug 09 '14 at 14:39

0 Answers0