Question

PLEASE HELP W PRECALC QUESTION

Answer 1

Answer:
`2tan(theta)`

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:

`√(x^2-4) =√((2sec(theta)^2-4) =√(4sec^2(theta)-4)`

Factor out a 4 to get:

`√(4(sec^2(theta)-1))`

Take the square root of 4 to get a 2 on the outside of the radical:

`√(4(sec^2(theta)-1))=2√(sec^2(theta)-1)`

Use the trig identity below to convert the secant to tangent:

`tan^2(theta)+1=sec^2(theta)`

`tan^2(theta)=sec^2(theta)-1`

Therefore our expression becomes:

`2√(sec^2(theta)-1)=2√(tan^2(theta))`

The square root of tangent is simply tangent:

`2√(tan^2(theta)) =2tan(theta)`