Answer: [tex]2tan(theta)[/tex]
Step-by-step explanation:
I know this is pre-calculus, but I want to let you know what this is actually used for in real calculus. This is a method of integration called "u-substitution" which tranforms the function of 'x' into a function of 'theta'. If we allow the value of 'x' to equal 2sec(theta), we can plug this into the function to get:
[tex]\sqrt{x^2-4} =\sqrt{(2sec(theta)^2-4} =\sqrt{4sec^2(theta)-4}[/tex]
Factor out a 4 to get:
[tex]\sqrt{4(sec^2(theta)-1)}[/tex]
Take the square root of 4 to get a 2 on the outside of the radical:
[tex]\sqrt{4(sec^2(theta)-1)}=2\sqrt{sec^2(theta)-1}[/tex]
Use the trig identity below to convert the secant to tangent:
[tex]tan^2(theta)+1=sec^2(theta)[/tex]
[tex]tan^2(theta)=sec^2(theta)-1[/tex]
Therefore our expression becomes:
[tex]2\sqrt{sec^2(theta)-1}=2\sqrt{tan^2(theta)}[/tex]
The square root of tangent is simply tangent:
[tex]2\sqrt{tan^2(theta)} =2tan(theta)[/tex]
