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$\cot^{-1}x = \tan^{-1}\dfrac{1}{x}$, $\forall$ $x\in R -$ {$0$}

Say I need to check if its true or not.

METHOD 1

Putting $\cot^{-1}x = \pi/2- \tan^{-1}x$

$$\Rightarrow \pi/2- \tan^{-1}x = \tan^{-1}\dfrac{1}{x}$$

$$\Rightarrow\pi/2 = \tan^{-1}\dfrac{1}{x}+ \tan^{-1}x$$

Using identity $\tan^{-1}x+\tan^{-1}y= \tan^{-1}\dfrac{x+y}{1-xy}$, we get

$$\Rightarrow\pi/2 = \tan^{-1}\dfrac{\dfrac{1}{x}+x}{1-1}$$

$$\Rightarrow\pi/2 = \tan^{-1}\infty$$

So we can say that the statement is true.

METHOD 2

Let $\cot^{-1}x=\theta$, where $\theta$ is an angle of a triangle. So $\cot\theta =x$.

But $\cot \theta = \dfrac{1}{\tan \theta}$

So $\dfrac{1}{\tan \theta}=x$

$\Rightarrow \tan \theta = \dfrac{1}{x}$

$\Rightarrow \tan^{-1}\dfrac{1}{x} = \theta $

$\Rightarrow \tan^{-1}\dfrac{1}{x} = \cot^{-1}x$

So the statement is true.

But the true solution is in this question through which we can clearly say that the above statement is false. I know I might have made every possible mistake in these two methods. I just want to know what are my mistakes and wrong assumptions.

Community
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manshu
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    Your first attempt is a proof in the wrong direction and it ends up with a nonsensical statement. So it cannot be correct. – Hans Engler Jan 30 '16 at 14:47
  • @HansEngler that's what I am asking. Where did I go wrong? Which one was the wrong direction? – manshu Jan 30 '16 at 14:49

2 Answers2

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Your first method only shows that by starting with a false hypothesis you can reach a true conclusion.

Your second method stumbles at the step $\tan\theta={1\over x}\implies\theta=\tan^{-1}{1\over x}$. There are many angles whose tangent is $1/x$; the inverse tangent function picks out a specific one, which may or may not be the angle you have in mind.

The easiest way to see that $\cot^{-1}x$ is not always equal to $\tan^{-1}(1/x)$ is to note that the inverse cotangent is always positive, while the inverse tangent function is negative for $x\lt0$.

Barry Cipra
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  • Which one was the wrong hypothesis in the first method? – manshu Jan 30 '16 at 15:06
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    @manshu, it appears you began by assuming the conclusion when you went from $\pi/2-\tan^{-1}x=\cot^{-1}x$ to $\pi/2-\tan^{-1}x=\tan^{-1}{1\over x}$. – Barry Cipra Jan 30 '16 at 15:22
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The statement "$\cot^{-1}x = \tan^{-1}(1/x)$ for all real $x > 0$" is true, and careful inspection shows your second proof works in this case. To show "$\cot^{-1}x = \tan^{-1}(1/x)$ for all real $x \neq 0$" is false, it suffices to find a single counterexample. A simple choice is $x = -1$; tracing through each argument with that specific value will show where your reasoning goes astray.

Andrew D. Hwang
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