Q.Which of the following is the principal value branch of $\cos^{-1}x$?
(A) $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$
(B) $(0,\pi)$
(C) $[0,\pi]$
(D) $(0,\pi)-\left\{\frac{\pi}{2}\right\}$

The key idea is that the principal value branch of $\cos^{-1}x$ is chosen so that the function is one-to-one and covers the full range of cosine values.

• Cosine is decreasing and one-to-one on $[0,\pi]$, and its range there is $[-1,1]$.
• The principal value of $\cos^{-1}x$ is defined as the unique angle $\theta$ in $[0,\pi]$ such that $\cos\theta = x$. …

Why the principal value branch matters

Inverse trigonometric functions are tricky because the original trigonometric functions are periodic — they are not one-to-one over their entire domain. If we tried to invert $\cos x$ without restriction, we’d get infinitely many angles for a single $x$, which is useless as a function. So we restrict the range (the output) to a specific interval where $\cos x$ is one-to-one and covers all possible values of $x$ in $[-1,1]$. That restricted range is called the principal value branch.

For $\cos^{-1}x$, we need an interval where $\cos \theta$ is:

• One-to-one (passes the horizontal line test)
• Onto $[-1,1]$ (every $x$ in $[-1,1]$ appears exactly once)

The natural choice is $[0,\pi]$, because on this interval $\cos \theta$ decreases strictly from $1$ to $-1$, covering every value exactly once.

Step-by-step reasoning

1. Understand what the question is asking

   The principal value branch of $\cos^{-1}x$ is the set of possible output angles $\theta$ such that $\theta = \cos^{-1}x$. We are given four candidate intervals and must pick the correct one.

2. Recall the definition

   For $x \in [-1,1]$, $\cos^{-1}x$ is defined as the unique angle $\theta$ in the principal value branch satisfying $\cos \theta = x$. The standard convention (used in NCERT and all Indian board exams) is:

   $\cos^{-1}x \in [0,\pi]$

3. Check each option against the requirement

   • Option (A): $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ — On this open interval, $\cos \theta$ is positive (since $\cos$ is positive in the first quadrant and negative in the second, but here we only have first quadrant). It never gives negative $x$ values, so it cannot cover $[-1,1]$. Also, $\cos \theta$ is not one-to-one on this interval? Actually it is one-to-one, but the range of $\cos$ on $(-\pi/2,\pi/2)$ is $(0,1]$, not $[-1,1]$. So this fails.
   • Option (B): $(0,\pi)$ — This open interval excludes the endpoints $0$ and $\pi$. At $\theta=0$, $\cos 0 = 1$; at $\theta=\pi$, $\cos\pi = -1$. If we exclude these endpoints, then $x=1$ and $x=-1$ would have no corresponding angle, so the function would not be defined for the full domain $[-1,1]$. Hence this is incomplete. …

Method: Identifying the principal value branch of an inverse trig function

A "which interval is the principal branch" question is answered not by memory alone but by the two conditions every principal branch must satisfy. Use them to test each option.

Steps

Step 1: State the two defining conditions.

The principal value branch is the interval of output angles on which the parent trig function is (a) one-to-one (strictly monotonic) and (b) onto its full range (hits every attainable value exactly once). Any interval failing either test cannot be the branch.

Step 2: Match the parent function's behaviour.

For $\cos^{-1}x$, look at where $\cos\theta$ is strictly monotonic and sweeps $[-1,1]$ once. On $[0,\pi]$, $\cos\theta$ decreases from $1$ to $-1$ — one-to-one and onto. Symmetric intervals like $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ fail because $\cos$ is two-to-one there.

Step 3: Check endpoints — open vs closed. …

Common Mistakes

Mistake 1: Choosing $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ — the $\sin^{-1}$ range — for $\cos^{-1}$.

Why it's wrong: on $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ cosine is two-to-one and never negative, so it cannot cover $[-1,1]$ one-to-one. Correct approach: $\cos^{-1}$ uses $[0,\pi]$, where cosine decreases from $1$ to $-1$.

Mistake 2: Picking an open interval $(0,\pi)$. …