Finding the range of a trigonometric function y = 1/(sinx*cosx)

[Nikolas111]

Hello everyone.

I found another example that I couldn't solve and after asking my teacher she couldn't solve it either, the example is to find the range of y=1/(sinx*cosx) I analyzed the solution by means of application Symbolab but it just looks way too long for counting without calculator and having like 3 minutes for solving it . So my question is how to solve this type of question or what's your strategy when you encounter this type of example.

Thank you in advance for everyone

 

[HallsofIvy]

It think the trig identity \(\displaystyle sin(2x)= 2 sin(x)cos(x)\) will help!

\(\displaystyle \frac{1}{sin(x)cos(x)}= \frac{2}{sin(2x)}\)

Of course sin(2x) goes from -1 to 1. Since we can't divide by 0, the domain has two separated parts, \(\displaystyle [1,\infty)\) and \(\displaystyle (-\infty, -1]\).

 

[Nikolas111]

Hello and thank you for your respond

Well that's the reason, I thought the same answer was correct sadly it was not the correct answer, the correct answer was (-infinity;2) disjunction (2;+infinity) so I don't know how to solve it .
Excuse me for writing symbols by names but I can't type those symbols on my keyboard.

 

[HallsofIvy]

I made a couple of misprints before. It is correct that this is \(\displaystyle \frac{2}{sin(2x)} \) and that sin(2x) has values from -1 to 1. From that it follows that the range (not "domain") of 1/(sin(x)cos(x)) is 2/(-1)= -2 to negative infinity and 2/(1)= 2 to I infinity.

 

[Nikolas111]

Alright Thank you very much for your help