# Given the function: f : [ &#x2212;<!-- − --> &#x03C0;<!-- π --> 2 </

Given the function:
$f:\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\to \mathbb{R}$
$f\left(x\right)=\frac{\mathrm{cos}x}{\mathrm{cos}\left(x-a\right)}$ for some $a\in \mathbb{R}$
Is it possible to convert it to some kind of translated / otherwise transformed tanx or cotx?
You can still ask an expert for help

## Want to know more about Trigonometry?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Crevani9a
Yes it is. You have just to sum ±a in teh argument of the upper cosine
$f\left(x\right)=\frac{\mathrm{cos}\left(x-a+a\right)}{\mathrm{cos}\left(x-a\right)}$
and use the expression for the sum of angles
$f\left(x\right)=\frac{\mathrm{cos}\left(x-a\right)\mathrm{cos}\left(a\right)-\mathrm{sin}\left(x-a\right)\mathrm{sin}\left(a\right)}{\mathrm{cos}\left(x-a\right)}=\mathrm{cos}\left(a\right)-\mathrm{tan}\left(x-a\right)\mathrm{sin}\left(a\right)$
###### Not exactly what you’re looking for?
res2bfitjq
You could also do the following.
$f\left(x\right)=\frac{\mathrm{cos}x}{\mathrm{cos}\left(x-a\right)}=\frac{1}{\frac{\mathrm{cos}\left(x-a\right)}{cos\left(x\right)}}=\frac{1}{\frac{\mathrm{cos}\left(x\right)\mathrm{cos}\left(a\right)+\mathrm{sin}\left(x\right)\mathrm{sin}\left(a\right)}{cos\left(x\right)}}=\frac{1}{\mathrm{cos}\left(a\right)+\mathrm{sin}\left(a\right)\mathrm{tan}\left(x\right)}$