Q.The value of $\cot^{-1}(-x)$ for all $x\in R$ in terms of $\cot^{-1}x$ is __________.

The key idea is the odd-like property of $\cot^{-1}$ under a sign change, but note that the range of $\cot^{-1}$ is $(0, \pi)$, so the relationship involves a shift.

Step 1: Recall the definition: $\cot^{-1}x \in (0, \pi)$ for all real $x$.

Step 2: For $x > 0$, $\cot^{-1}(-x) = \pi - \cot^{-1}x$, because $\cot(\pi - \theta) = -\cot\theta$ and $\pi - \theta$ stays in $(0, \pi)$.

Step 3: For $x < 0$, the same formula holds: if $x$ is negative, $-x$ is positive, and the relation still gives $\cot^{-1}(-x) = \pi - \cot^{-1}x$. …

Why this approach works

The inverse cotangent function, $\cot^{-1}x$, is defined to give an angle in the principal value range $(0, \pi)$. This is a crucial choice: unlike $\tan^{-1}x$ which lives in $(-\pi/2, \pi/2)$, $\cot^{-1}x$ lives strictly between $0$ and $\pi$, never touching $0$ or $\pi$ themselves.

Now, what happens when we put a negative input, $-x$? The angle $\cot^{-1}(-x)$ must also lie in $(0, \pi)$. But the cotangent of an angle $\theta$ in $(0, \pi)$ is negative only when $\theta$ is in $(\pi/2, \pi)$. So $\cot^{-1}(-x)$ always lands in the second quadrant — between $\pi/2$ and $\pi$.

Meanwhile, $\cot^{-1}x$ (for positive $x$) lies in $(0, \pi/2)$, and for negative $x$ it lies in $(\pi/2, \pi)$. The relationship we need must connect these two angles cleanly.

The trick is to use the identity $\cot(\pi - \theta) = -\cot\theta$. This tells us: if $\cot^{-1}x = \theta$, then $\cot(\pi - \theta) = -x$. So $\pi - \theta$ is an angle whose cotangent is $-x$, and it lies in $(0, \pi)$ as long as $\theta$ does. That makes it the principal value — exactly $\cot^{-1}(-x)$.

Let's verify this step by step.

1. Set up the notation.

   Let $\theta = \cot^{-1}x$. By definition, $\theta \in (0, \pi)$ and $\cot\theta = x$.

2. Consider $\pi - \theta$.

   Since $\theta \in (0, \pi)$, we have $\pi - \theta \in (0, \pi)$ as well.

   Using the identity $\cot(\pi - \theta) = -\cot\theta$, we get:

3. Check the range.

   $\pi - \theta$ is in $(0, \pi)$, which is exactly the principal value range of $\cot^{-1}$.

   Therefore, $\pi - \theta$ is a valid candidate for $\cot^{-1}(-x)$.

4. Conclude the equality. …

Method: Derive a negative-argument identity from the principal range

Steps

Step 1: Set $\theta=\cot^{-1}x$, so $\cot\theta=x$ and $\theta\in(0,\pi)$.

Step 2: Find an angle with cotangent $-x$ inside $(0,\pi)$.

Use $\cot(\pi-\theta)=-\cot\theta=-x$, and note $\pi-\theta\in(0,\pi)$ whenever $\theta$ is. …

Common Mistakes

Mistake 1: Writing $\cot^{-1}(-x)=-\cot^{-1}x$.

Why it's wrong: it copies the odd-function rule of $\tan^{-1}$, but $-\cot^{-1}x$ is negative for $x>0$, which falls outside the range $(0,\pi)$. Correct approach: use $\pi-\cot^{-1}x$.

Mistake 2: Assuming $\cot^{-1}$ has range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$. …