Q.If $\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0$, then value of $x$ is __________.

Concept: Inverse trigonometric identities — specifically, $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ for all real $x$, and the condition $\cos\theta = 0$ means $\theta = (2n+1)\frac{\pi}{2}$.

Step 1:

We are given $\cos(\tan^{-1}x+\cot^{-1}\sqrt3)=0$.

Since $\cos\theta = 0$ implies $\theta = \frac{\pi}{2} + n\pi$, we have:

Step 2:

Recall $\cot^{-1}\sqrt3 = \frac{\pi}{6}$ (since $\cot\frac{\pi}{6} = \sqrt3$).

So the equation becomes:

$\tan^{-1}x + \frac{\pi}{6} = \frac{\pi}{2} + n\pi.$ …

The problem asks for $x$ when $\cos(\tan^{-1}x + \cot^{-1}\sqrt{3}) = 0$. This is a composite trigonometric equation. The core idea: $\cos\theta = 0$ means $\theta$ is an odd multiple of $\frac{\pi}{2}$, but here the angles are principal values (since $\tan^{-1}$ and $\cot^{-1}$ are defined with specific ranges), so we can directly set the sum equal to $\frac{\pi}{2}$.

Why? Because $\tan^{-1}x$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ and $\cot^{-1}\sqrt{3}$ lies in $(0, \pi)$. Their sum could be many things, but for the cosine to be zero, the sum must be $\frac{\pi}{2}$ (or $-\frac{\pi}{2}$, but that’s impossible here — check ranges). So we solve:

1. Set the argument equal to $\frac{\pi}{2}$ Since $\cos\theta = 0$ implies $\theta = \frac{\pi}{2} + n\pi$, but the principal values restrict us. $\cot^{-1}\sqrt{3}$ is a positive acute angle: $\cot^{-1}\sqrt{3} = \frac{\pi}{6}$ (since $\cot\frac{\pi}{6} = \sqrt{3}$). And $\tan^{-1}x$ is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. Their sum can only be $\frac{\pi}{2}$ (not $-\frac{\pi}{2}$ or $\frac{3\pi}{2}$). So: $\tan^{-1}x + \frac{\pi}{6} = \frac{\pi}{2}$ …

Method: Solving an equation wrapped in an outer trig function

Use this for equations of the form $\cos(\text{inverse-trig expression})=0$ (or $\sin(\cdots)=0$, etc.).

Steps

Step 1: Undo the outer function.

$\cos\theta=0$ means $\theta$ is an odd multiple of $\tfrac{\pi}{2}$: write $\theta=\tfrac{\pi}{2}+n\pi$.

Step 2: Pin down the reachable value using principal ranges. …

Common Mistakes

Mistake 1: Taking $\cot^{-1}\sqrt3=\frac{\pi}{3}$.

Why it's wrong: $\frac{\pi}{3}$ is $\tan^{-1}\sqrt3$; since $\cot\frac{\pi}{6}=\sqrt3$, we have $\cot^{-1}\sqrt3=\frac{\pi}{6}$. Correct approach: check which ratio equals $\sqrt3$ before quoting the angle.

Mistake 2: Keeping every odd multiple of $\frac{\pi}{2}$. …