Q.Find the principal value of the following: $\tan^{-1} \left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$, $0 < x < \pi$

The key idea is to simplify the inner expression using a half-angle identity, then apply the principal range of $\tan^{-1}$.

Step 1: Use the identities $1 - \cos x = 2\sin^2\frac{x}{2}$ and $1 + \cos x = 2\cos^2\frac{x}{2}$.

Step 2: Substitute:

Step 3: Since $0 < x < \pi$, we have $0 < \frac{x}{2} < \frac{\pi}{2}$, so $\tan\frac{x}{2} > 0$. Hence $\left|\tan\frac{x}{2}\right| = \tan\frac{x}{2}$.

Step 4: Therefore, …

We are asked to find the principal value of

The key here is to simplify the inside expression using trigonometric identities, then apply the definition of the principal value of the inverse tangent function. The range of $\tan^{-1}$ is $(-\pi/2, \pi/2)$, so we must ensure our final angle lies in that interval.

1. Simplify the fraction under the square root Recall the half-angle identities:

Therefore,

2. Take the square root

The absolute value is crucial — the square root always gives a non-negative result.

3. Determine the sign of $\tan(x/2)$ in the given interval

   Since $0 < x < \pi$, we have $0 < x/2 < \pi/2$. In this first quadrant, $\tan(x/2) > 0$.

   Hence $|\tan(x/2)| = \tan(x/2)$.

   So the original expression becomes

4. Apply the principal value The principal value branch of $\tan^{-1}$ is $(-\pi/2, \pi/2)$. Since $0 < x/2 < \pi/2$, the angle $x/2$ lies strictly inside this interval. …

Method: Simplifying an inverse trig expression with half-angle identities

Use this when the argument mixes $1 \pm \cos x$ (or $1 \pm \sin x$) under a root.

Steps

Step 1: Replace with half-angle forms

Apply $1 - \cos x = 2\sin^2\frac{x}{2}$ and $1 + \cos x = 2\cos^2\frac{x}{2}$, so the ratio collapses to $\tan^2\frac{x}{2}$.

Step 2: Take the root as an absolute value …

Common Mistakes

Mistake 1: Dropping the absolute value after the square root

Why it's wrong: $\sqrt{\tan^2\frac{x}{2}} = \left|\tan\frac{x}{2}\right|$; ignoring it gives the wrong sign when $\frac{x}{2}$ is obtuse. Correct approach: here $0 < x < \pi$ makes $\frac{x}{2}$ acute, so $\tan\frac{x}{2} > 0$ and the bars drop safely.

Mistake 2: Cancelling $\tan^{-1}\left(\tan\frac{x}{2}\right)$ to $\frac{x}{2}$ without checking the branch …