We know the following relationship:
[tex]csc(\theta)=\frac{1}{sin(\theta)}[/tex]
The domain of a function are the inputs of the function, that is, a function [tex]f[/tex] is a relation that assigns to each element [tex]x[/tex] in the set A exactly one element in the set B. The set A is the domain (or set of inputs) of the function and the set B contains the range (or set of outputs).Then applying this concept to our function [tex]csc(\theta)[/tex] we can write its domain as follows:
1. Domain of validity for [tex]csc(\theta)[/tex]:
[tex]D: \{\theta \in R/ sin(\theta) \neq 0 \} \\ In words: All \ \theta \ that \ are \ real \ values \ except \ those \ that \ makes \ sin(\theta)=0[/tex]
When:
[tex]sin(\theta)=0[/tex]?
when:
[tex]\theta=..., -2\pi,-\pi,0,\pi,\2pi,3pi,...,k\pi[/tex]
where k is an integer either positive or negative. That is:
[tex]sin(k\theta)=0 \ for \ k=...,-2,-1,0,1,2,3,...[/tex]
To match this with the choices above, the answer is:
"All real numbers except multiples of [tex]\pi[/tex]"
2. which identity is not used in the proof of the identity [tex]1+cot^{2}(\theta)=csc^{2}(\theta)[/tex]:
This identity can proved as follows:
[tex]sin^2{\theta}+cos^{2}(\theta)=1 \ Dividing \ by \ sin^{2}(\theta) \\ \\ \therefore \frac{sin^2{\theta}}{sin^{2}(\theta)}+\frac{cos^{2}(\theta)}{sin^{2}(\theta)}=\frac{1}{sin^{2}(\theta)} \\ \\ \therefore 1+cot^{2}(\theta)=csc^{2}(\theta)[/tex]
The identity that is not used is as established in the statement above:
"1 +cos squared theta over sin squared theta= csc2theta"
Written in mathematical language as follows:
[tex]\frac{1+cos^{2}(\theta)}{sin^{2}(\theta)}=csc^{2}(\theta)[/tex]
