Example 4: Simplest form of $\tan^{-1}\frac{\cos x}{1-\sin x}$

Basics you need

Solution

1. Check the fraction is defined, and locate $\frac{x}{2}$

   The fraction needs $1 - \sin x \ne 0$⁠. Now $\sin x = 1$ only at $x = \frac{\pi}{2} + 2n\pi$⁠. Two consecutive such values are $-\frac{3\pi}{2}$ (with $n=-1$⁠) and $\frac{\pi}{2}$ (with $n=0$⁠); they are $2\pi$ apart, so no other one lies between them, and they are exactly the endpoints of our interval, which are excluded. Since $\sin x\le1$ always, on $-\frac{3\pi}{2} < x < \frac{\pi}{2}$ we have $1-\sin x > 0$ and the fraction is defined. Halving the interval:

   $$-\frac{3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}.$$
   The interval in the question is not decoration. It is the gap between two consecutive points where $1-\sin x = 0$⁠, and, as Step 5 shows, it is exactly the set of $x$ for which the final answer is correct. Knowing where $\frac{x}{2}$ lies is what lets us justify the cancellations below.
   $\tan^{-1}$ accepts every real number, so once the fraction is defined, $f(x)$ is defined. Lesson: principal values

2. Rewrite the top and bottom with half angles

   Write $c = \cos\frac{x}{2}$ and $s = \sin\frac{x}{2}$ to keep the algebra short. By Basic 1,

   $$\begin{aligned} \cos x &= c^2 - s^2 = (c+s)(c-s), \\ 1 - \sin x &= (c^2+s^2) - 2sc = (c-s)^2. \end{aligned}$$

   So

   $$\frac{\cos x}{1-\sin x} = \frac{(c+s)(c-s)}{(c-s)^2}.$$
   This matches the textbook's first two lines. The trick in the denominator is to write the number 1 as $\cos^2\frac{x}{2}+\sin^2\frac{x}{2}$⁠. Then $1 - \sin x$ has the shape $a^2+b^2-2ab$⁠, a perfect square. The numerator, a difference of two squares, then shares the factor $c - s$ with it.
   $\cos2\theta = \cos^2\theta-\sin^2\theta$ and $\sin2\theta = 2\sin\theta\cos\theta$⁠, with $\theta = \frac{x}{2}$⁠.

3. Cancel $c - s$⁠, after checking it is not zero

   By the cosine addition formula,

   $$\cos\left(\frac{x}{2}+\frac{\pi}{4}\right) = \frac{1}{\sqrt2}\left(\cos\frac{x}{2} - \sin\frac{x}{2}\right),$$

   so

   $$c - s = \sqrt2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right).$$

   By Step 1, $\frac{x}{2}+\frac{\pi}{4}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠, where cosine is positive. So $c - s > 0$⁠, it is not zero, and we may cancel one factor:

   $$\begin{aligned} \frac{\cos x}{1-\sin x} &= \frac{c+s}{c-s} \\ &= \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}}. \end{aligned}$$
   The textbook cancels without comment. Cancelling a factor that could be 0 would be dividing by zero. Here $c - s = 0$ would need $\tan\frac{x}{2} = 1$⁠, i.e. $\frac{x}{2} = \frac{\pi}{4}+n\pi$⁠; none of these is in $\left(-\frac{3\pi}{4}, \frac{\pi}{4}\right)$⁠, confirming the cancellation is safe.
   $\cos(A+B) = \cos A\cos B - \sin A\sin B$ with $B = \frac{\pi}{4}$⁠, where $\cos\frac{\pi}{4} = \sin\frac{\pi}{4} = \frac{1}{\sqrt2}$⁠.

4. Recognise $\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)$

   The textbook's route. Divide top and bottom by $\cos\frac{x}{2}$⁠:

   $$\begin{aligned} \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}} &= \frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}} \\ &= \tan\left(\frac{\pi}{4}+\frac{x}{2}\right), \end{aligned}$$

   the last step by the addition formula with $\tan\frac{\pi}{4} = 1$ (Basic 3). This division needs $\cos\frac{x}{2}\ne0$⁠. In our interval $\frac{x}{2} = -\frac{\pi}{2}$ is possible, at $x = -\pi$, where $\tan\frac{x}{2}$ is undefined. So the textbook's route misses that one point.

   A route with no gap. By the sine addition formula,

   $$\sin\left(\frac{x}{2}+\frac{\pi}{4}\right) = \frac{1}{\sqrt2}\left(\sin\frac{x}{2}+\cos\frac{x}{2}\right),$$

   so $c + s = \sqrt2\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)$⁠. With Step 3,

   $$\begin{aligned} \frac{c+s}{c-s} &= \frac{\sqrt2\sin\left(\frac{x}{2}+\frac{\pi}{4}\right)}{\sqrt2\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)} \\ &= \tan\left(\frac{\pi}{4}+\frac{x}{2}\right). \end{aligned}$$

   This only divides by $\cos\left(\frac{x}{2}+\frac{\pi}{4}\right)$⁠, which Step 3 showed is positive for every $x$ in the interval, including $x = -\pi$⁠. Directly at $x=-\pi$⁠: $\frac{\cos(-\pi)}{1-\sin(-\pi)} = \frac{-1}{1} = -1$⁠, and $\tan\left(\frac{\pi}{4}-\frac{\pi}{2}\right) = \tan\left(-\frac{\pi}{4}\right) = -1$⁠. It agrees.

   Both routes reach the same expression. The textbook's is the one to recognise in an exam, since $\frac{1+\tan B}{1-\tan B}$ is a standard pattern; the second shows the result holds at every point of the interval with no exception.
   $\sin(A+B) = \sin A\cos B + \cos A\sin B$⁠; $\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B}$⁠.

5. Cancel $\tan^{-1}(\tan\,\cdot)$⁠, after checking the angle

   So far, for every $x$ in the interval,

   $$f(x) = \tan^{-1}\left[\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right].$$

   Add $\frac{\pi}{4}$ to each part of $-\frac{3\pi}{4} < \frac{x}{2} < \frac{\pi}{4}$⁠:

   $$-\frac{\pi}{2} < \frac{\pi}{4}+\frac{x}{2} < \frac{\pi}{2}.$$

   The angle is in the principal range of $\tan^{-1}$⁠, so by Basic 4

   $$\tan^{-1}\left(\frac{\cos x}{1-\sin x}\right) = \frac{\pi}{4}+\frac{x}{2}.$$
   This is the step that explains the interval $-\frac{3\pi}{2} < x < \frac{\pi}{2}$⁠: it is precisely the set of $x$ for which $\frac{\pi}{4}+\frac{x}{2}$ lands in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$⁠. For $x$ just beyond $\frac{\pi}{2}$⁠, the angle $\frac{\pi}{4}+\frac{x}{2}$ exceeds $\frac{\pi}{2}$⁠, and $\tan^{-1}$ returns $\frac{\pi}{4}+\frac{x}{2}-\pi$ instead (see the check below).
   $\tan^{-1}(\tan t) = t$ for $-\frac{\pi}{2} < t < \frac{\pi}{2}$ (Section 2.3, which states the rule for $\sin^{-1}$ and adds that "similar results follow for remaining trigonometric functions"). Lesson: substitutions and simplest form

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