# Math (Double checking trig identity)

Since sec^2θ - 1 = tan^2θ

I know this is trivial, but I want to make sure I'm doing this right before I apply it to the integral I'm trying to solve...

If I have some constant a,

(a^2secθ)^2 - (a^2)^2

If I wanted to change this to tan would it be:

a^4tan^4θ?

Any help is greatly appreciated!

1. not at all.

(a^2secθ)^2 - (a^2)^2
= a^4 (sec^4θ - 1)
= a^4 (sec^2θ-1)(sec^2θ+1)
= a^4 tan^2θ (sec^2θ+1)

posted by Steve
2. Nope. my bad. Still you were also wrong

(a^2secθ)^2 - (a^2)^2
= a^4 (sec^2θ-1)
= a^4 tan^2θ

posted by Steve
3. So tan^2θ is where I messed up, thanks Steve!

posted by Ray

## Similar Questions

1. ### Integration

Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? thanks. = ¡ì sec x d tan x = sec x tan x - ¡ì tan x d sec x = sec x tan x - ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx - ¡ì
2. ### calculus (check my work please)

Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)-1) dx ∫ tan(x)sec(x)[sec^4(x)-sec^2(x)]

2- given the curve is described by the equation r=3cos ¥è, find the angle that the tangent line makes with the radius vector when ¥è=120¨¬. A. 30¨¬ B. 45¨¬ C. 60¨¬ D. 90¨¬ not sure A or D 2.) which of the following
4. ### calculus

find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan
5. ### Calculus AP

I'm doing trigonometric integrals i wanted to know im doing step is my answer right? ∫ tan^3 (2x) sec^5(2x) dx =∫ tan^2(2x) sec^4(2x) tan*sec(2x) dx =∫ (sec^2(2x)-1)sec^4 tan*sec(2x) dx let u=sec x, du= 1/2 tan*sec(2x) dx
6. ### calculus

So I am suppose to evaulate this problem y=tan^4(2x) and I am confused. my friend did this : 3 tan ^4 (2x) d sec^ 2x (2x)= 6 tan ^4 (2x) d sec^2 (2x) She says it's right but what confuses me is she deriving the 4 and made it a
7. ### Calculus

could anybody please explain how sec x tan x - ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx - ¡ì sec^3(x) dx What I don't understand about your question is what is ¡ì ? i just want to know if those two equations are
8. ### more trig.... how fun!!!!

if you can't help me with my first question hopw you can help me with this one. sec(-x)/csc(-x)=tan(x) thanx to anyone who can help From the definition of the sec and csc functions, and the tan function, sec(-x)/csc(-x) =