Answer:
[tex]\sec(x)=\frac{1}{\sqrt{1-m^2}}[/tex]
Step-by-step explanation:
We know that:
[tex]\sin(-x)=-m[/tex]
First, since sine is an odd function, we can move the negative outside:
[tex]=-\sin(x)=-m[/tex]
Divide both sides by -1:
[tex]\sin(x)=m[/tex]
We will now use the Pythagorean Identity:
[tex]\cos^2(x)+\sin^2(x)=1[/tex]
Substitute m for sine:
[tex]\cos^2(x)+m^2=1[/tex]
Solve for cosine:
[tex]\cos^2(x)=1-m^2[/tex]
Take the square root of both sides:
[tex]\cos(x)=\pm\sqrt{1-m^2}[/tex]
Since x is an acute angle, cosine will always be positive. Thus:
[tex]\cos(x)=\sqrt{1-m^2}[/tex]
Take the reciprocal of both sides. Hence:
[tex]\frac{1}{\cos(x)}=\sec(x)=\frac{1}{\sqrt{1-m^2}}[/tex]
