Q.Find the principal value of the following: $\tan^{-1} \left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$, $-\frac{\pi}{4} < x < \frac{3\pi}{4}$

Concept: Inverse Trigonometric Graphs — The principal value branch of $\tan^{-1}$ lies in $(-\pi/2, \pi/2)$. We simplify the argument to a single tangent so the inverse can be read directly.

Step 1: Divide numerator and denominator by $\cos x$ (valid since $\cos x \neq 0$ in the given interval except at $x = \pi/2$, but the expression is still defined by limit):

Step 2: Recognise the formula $\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$. Hence:

$\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right) = \tan^{-1}\left[\tan\left(\frac{\pi}{4} - x\right)\right].$ …

We need the principal value of $\tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$ for $x$ in the interval $(-\pi/4, 3\pi/4)$. The "principal value" of an inverse trigonometric function means the unique angle in its principal branch — for $\tan^{-1}$, that's $(-\pi/2, \pi/2)$. So our job is to simplify the expression inside until it matches $\tan(\text{something})$, then check that "something" lies in $(-\pi/2, \pi/2)$ for the given $x$ range.

The trick is to see the numerator and denominator as a disguised tangent subtraction formula. Divide numerator and denominator by $\cos x$ (valid since $\cos x \neq 0$ in most of the interval — we'll check the edge later).

1. Rewrite in terms of $\tan x$

because $\frac{\sin x}{\cos x} = \tan x$.

2. Recognise the tangent subtraction formula Recall: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$. If we set $A = \frac{\pi}{4}$ and $B = x$, then $\tan\frac{\pi}{4} = 1$, so

Exactly our expression! So

3. Apply the inverse tangent Therefore,

But $\tan^{-1}(\tan \theta)$ equals $\theta$ only when $\theta$ lies in the principal branch $(-\pi/2, \pi/2)$. Otherwise, we need to adjust by adding or subtracting $\pi$.

4. Check the range of $\frac{\pi}{4} - x$ Given $-\frac{\pi}{4} < x < \frac{3\pi}{4}$, multiply by $-1$ (reversing inequalities):

Then add $\frac{\pi}{4}$:

$-\frac{\pi}{2} < \frac{\pi}{4} - x < \frac{\pi}{2}$ …

Method: Simplifying an inverse tangent by dividing through

Use this when the $\tan^{-1}$ argument is a ratio of the form $\frac{\cos x \pm \sin x}{\cos x \mp \sin x}$.

Steps

Step 1: Divide numerator and denominator by $\cos x$

This turns the ratio into $\frac{1 - \tan x}{1 + \tan x}$ (or the sign-flipped version), a tangent-subtraction pattern.

Step 2: Recognise the compound-angle form

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

Common Mistakes

Mistake 1: Not spotting the $\tan\left(\frac{\pi}{4} - x\right)$ pattern

Why it's wrong: leaving the ratio as $\frac{\cos x - \sin x}{\cos x + \sin x}$ blocks any simplification. Correct approach: divide by $\cos x$ to reveal $\frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right)$.

Mistake 2: Cancelling to $\frac{\pi}{4} - x$ without checking the branch …