Q.Express $\tan^{-1}\left(\dfrac{\cos x}{1-\sin x}\right)$, $-\dfrac{3\pi}{2} < x < \dfrac{\pi}{2}$ in the simplest form.

Idea: Write $\cos x$ and $1-\sin x$ using half-angles, cancel the common factor, and the fraction becomes a single tangent.

Step 1 — half-angle forms. Using $\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2}$ and $1-\sin x = \left(\cos\frac{x}{2}-\sin\frac{x}{2}\right)^2$:

Step 2 — divide by $\cos\frac{x}{2}$:

$\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}} = \tan\left(\frac{\pi}{4}+\frac{x}{2}\right)$ …

The trick with an expression like $\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right)$ is to turn the messy fraction inside into a single tangent. Once it looks like $\tan^{-1}(\tan\theta)$, the answer is just $\theta$ — provided $\theta$ sits inside the principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ of $\tan^{-1}$. So the plan is: simplify to a tangent, then check the domain.

Step 1 — Express in half-angles

Use the standard double-angle identities:

Then, since $\sin^2\tfrac{x}{2}+\cos^2\tfrac{x}{2}=1$,

Step 2 — Cancel the common factor

Factor the numerator as a difference of squares:

So

Step 3 — Turn it into a single tangent

Divide the top and bottom by $\cos\tfrac{x}{2}$:

This matches the tangent-addition formula with $\frac{\pi}{4}$, since $\tan\frac{\pi}{4}=1$:

With $\theta=\tfrac{x}{2}$,

Step 4 — Apply the inverse tangent (and check the domain) …

Method: Simplifying an inverse-trig expression via half-angle substitution

Use this when the argument of $\tan^{-1}$ (or another inverse) is a fraction in $\sin x$ and $\cos x$ that resists direct identities — the half-angle rewrite usually collapses it to a single tangent.

Steps

Step 1: Rewrite numerator and denominator in half-angles.

Replace $\cos x$ and $1\pm\sin x$ using

A perfect-square denominator is the sign this method will work.

Step 2: Factor and cancel to expose a $\dfrac{1+\tan}{1-\tan}$ shape.

After cancelling the common factor and dividing top and bottom by $\cos\tfrac{x}{2}$, aim for

$\frac{1+\tan\tfrac{x}{2}}{1-\tan\tfrac{x}{2}} = \tan\!\left(\tfrac{\pi}{4}+\tfrac{x}{2}\right),$ …

Common Mistakes

Mistake 1: Skipping the domain check before writing $\tan^{-1}(\tan\phi)=\phi$.

Why it's wrong: the cancellation is legal only when $\phi=\tfrac{\pi}{4}+\tfrac{x}{2}$ lies in $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$; outside it you must add or subtract $\pi$. Correct approach: feed the given $-\tfrac{3\pi}{2}<x<\tfrac{\pi}{2}$ through $\phi$ and verify $-\tfrac{\pi}{2}<\phi<\tfrac{\pi}{2}$ before removing the inverse.

Mistake 2: Mishandling the half-angle form of $1-\sin x$. …