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I was just wondering: We know that $\cot ^{-1}(\cot x) = x$. But what is the algebraic value of $\cot^{-1}(-\cot x)$? Is it $-x$?

I am just getting a little muddled up with my trigonometric identities. I would appreciate some clarification.

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3 Answers

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HINT

Remember that the cotangent function is odd:\begin{align*} \cot(x) = \frac{\cos(x)}{\sin(x)} \Longrightarrow \cot(-x) = \frac{\cos(-x)}{\sin(-x)} = -\frac{\cos(x)}{\sin(x)} = -\cot(x) \end{align*}

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$$ \cot^{-1}(\cot(x)) = x$$is not actually a trigonometric 'identity', it is a result of inverse function, $ f(f^{-1}(y)) = y $$$ \cot( \cot^{-1}(\cot(x)) ) = \cot(x) \implies \cot^{-1}(\cot(x)) = x $$

by @APC89's answer:$$ \cot( \cot^{-1}(-\cot(x)) ) = -\cot(x) = \cot(-x) $$so we have $$ \cot^{-1}(-\cot(x))= -x $$

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Yes, it is. Working backwards a bit, we can confirm this.
Remember that $\cos (-x)$ = $\cos (x)$ and $\sin (-x)$ = $-\sin (x)$.

$$\cot^{-1}(-\cot x) = \cot^{-1}\bigg(\frac{\cos (-x)}{\sin (-x)}\bigg) = \cot^{-1} (\cot (-x)) = -x$$

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