# Simplify $\tan^{-1}[(\cos x - \sin x)/(\cos x + \sin x)]$

Write the following functions in simplest form: $$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right), \quad 0<x<\pi$$

Please help me to solve this problem. I have been trying to solve this from last 3 hours. I can solve simple inverse trigonometric functions

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## migrated from mathematica.stackexchange.comAug 23 '12 at 13:05

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define simplest please. – Silvia Aug 23 '12 at 13:03
It would be nice if it had the form $\tan^{-1}\left(\frac{\sin(\cdots)}{\cos(\cdots)}\right)$, wouldn't it? See if you can use rewritings like this to make it so. – Henning Makholm Aug 23 '12 at 13:16

Let's multiply the numerator and denominator by $\ \cos\bigl(\frac {\pi}4\bigr)=\sin\bigl(\frac {\pi}4\bigr)\$ : $$\tan^{-1}\left(\frac{\sin\bigl(\frac {\pi}4\bigr)\cos(x)-\cos\bigl(\frac {\pi}4\bigr)\sin(x)}{\cos\bigl(\frac {\pi}4\bigr)\cos(x)+\sin\bigl(\frac {\pi}4\bigr)\sin(x)}\right),\quad 0<x<\pi$$

$$=\tan^{-1}\left(\frac{\sin\left(\frac {\pi}4-x\right)}{\cos\left(\frac {\pi}4-x\right)}\right),\quad 0<x<\pi$$ $$=\begin{cases} &\frac {\pi}4-x&\quad\text{if}\ \ 0 < x < \frac{3\pi}4\\ &\text{not defined}&\quad\text{if}\quad x = \frac{3\pi}4\\ &\frac {5\pi}4-x&\quad\text{if}\ \ \frac{3\pi}4< x < \pi\\ \end{cases}$$

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Let $\theta=\tan^{-1}(X)$ where $X=\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right)$ and $0<x<\pi$ ,we have $$\frac{1-X}{1+X}=\tan(x)$$ But $\tan(\theta)=X$ so, $\tan(x)=\frac{1-\tan(\theta)}{1+\tan(\theta)}=\tan(\frac{\pi}{4}-\theta)$. The rest is as Raymond concluded above.

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Nice $\checkmark\quad\ddot\smile$ – amWhy Mar 13 '13 at 1:00

$$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right),\quad 0<x<\pi$$

$=\tan^{-1}\left(\frac{1-\tan(x)}{1+\tan(x)}\right)$ diving the numerator and denominator by $\cos x$

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

$=\tan^{-1}\tan(\frac{\pi}{4}-x)$

$=n\pi+\frac{\pi}{4}-x$ where n is any integer.

The principal value which must lie in $[-\frac{\pi}{2},\frac{\pi}{2}]$, will be

$\frac{\pi}{4}-x$ when $\frac{\pi}{4}-x$ lies in that region i.e, $\frac{3\pi}{4} ≥ x > -\frac{\pi}{4}$

and $\pi +\frac{\pi}{4}-x$ elsewhere.

But $0<x<\pi$,

so, if $\frac{\pi}{2} ≥ x ≥ 0$ the principal value =$\frac{\pi}{4}-x$

and $\frac{5\pi}{4}-x$ elsewhere.

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