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

  • $\begingroup$ define simplest please. $\endgroup$ – Silvia Aug 23 '12 at 13:03
  • $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica 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)\ $ : $$\tan^{-1}\left(\frac{\sin\bigl(\frac {\pi}4\bigr)\cos(x)-\cos\bigl(\frac {\pi}4\bigr)\sin(x)}{\cos\bigl(\frac {\pi}4\bigr)\cos(x)+\sin\bigl(\frac {\pi}4\bigr)\sin(x)}\right),\quad 0<x<\pi$$

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


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.

  • $\begingroup$ Nice $\checkmark\quad\ddot\smile$ $\endgroup$ – 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 $\tan\frac{\pi}{4}=1$


$=n\pi+\frac{\pi}{4}-x$ 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 $\tan ^{-1}\dfrac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)+\sin \left(x\right)}$

We will divide in bracket with cosx to get it in tan form which will be easy for us to simplify

$\implies \tan ^{-1}\dfrac{\dfrac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)}}{\dfrac{\cos \left(x\right)-\sin \left(x\right)}{\cos \left(x\right)}}$

$\implies \tan ^{-1}\dfrac{1-\tan \left(x\right)}{1+\cos \left(x\right)}$

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 $\dfrac{\tan \left(1\right)-\tan \left(x\right)}{1-\tan \left(1\right)\tan \left(x\right)}$

So we get $\tan ^{-1} \left(\tan \left(\dfrac{π}{4}\right) + \tan \left(X\right)\right)$ i.e $\dfrac{π}{4-x}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.