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
Let's multiply the numerator and denominator by $\ \cos\bigl(\frac {\pi}4\bigr)=\sin\bigl(\frac {\pi}4\bigr)\ $ : $$\tan^{-1}\left(\frac{\sin\bigl(\frac {\pi}4\bigr)\cos(x)-\cos\bigl(\frac {\pi}4\bigr)\sin(x)}{\cos\bigl(\frac {\pi}4\bigr)\cos(x)+\sin\bigl(\frac {\pi}4\bigr)\sin(x)}\right),\quad 0<x<\pi$$
$$=\tan^{-1}\left(\frac{\sin\left(\frac {\pi}4-x\right)}{\cos\left(\frac {\pi}4-x\right)}\right),\quad 0<x<\pi$$ $$=\begin{cases} &\frac {\pi}4-x&\quad\text{if}\ \ 0 < x < \frac{3\pi}4\\ &\text{not defined}&\quad\text{if}\quad x = \frac{3\pi}4\\ &\frac {5\pi}4-x&\quad\text{if}\ \ \frac{3\pi}4< x < \pi\\ \end{cases} $$
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.
$$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right),\quad 0<x<\pi$$
$=\tan^{-1}\left(\frac{1-\tan(x)}{1+\tan(x)}\right)$ 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.
It is $tan^{-1}\dfrac{cos(x)-sin(x)}{cos(x)+sin(x)}$
We will divide in bracket with cosx to get it in tan form which will be easy for us to simplify
$\implies tan^{-1}\dfrac{\dfrac{cos(x)-sin(x)}{cos(x)}}{\dfrac{cos(x)-sin(x)}{cos(x)}}$
$\implies tan^{-1}\dfrac{1-tan(x)}{1+cos(x)}$
Now we know that $tan^{-1}x + tan^{-1}y= \dfrac{x-y}{1-xy}$ when $xy<1$
and we have the bracket in the same form as $\dfrac{tan(1)-tan(x)}{1-tan(1)tan(x)}$
So we get $tan^{-1} (tan (\dfrac{pi}{4}) + tan (X))$ i.e $\dfrac{pi}{4-x}$.
