Simplifying/Proving a Trigonometric Question

[BlackHole213]

Homework Statement

Prove:

secA(1+sinA) = (1+sinA+cosA)/(1-sinA+cosA)

Homework Equations

I have the following equations:

sin²x + cos²x = 1
1 + tan²x = sec²x
1 + cot²x = csc²x
sin(A+B) = sinAcosB + cosAsinB
sin(A-B) = sinAcosB - cosAsinB
cos(A+B) = cosAcosB - sinAsinB
cos(A-B) = cosAcosB + sinAsinB
sin(2A) = 2sinAcosA
cos(2A) = cos² - sin²A

These formulas are all I have, but as this is a bonus question, there might be other formulas which I am unaware of.

The Attempt at a Solution

I looked at https://www.physicsforums.com/showthread.php?t=423847 to see if I could get inspiration to achieve an answer. I've tried simplifying the left side to (1+sinA)/cosA and then I attempted to see if I could get the right hand side to equal that. I've multipled the RHS by (1+sinA+cosA)/(1+sinA+cosA), (1-sinA+cosA)/(1-sinA+cosA) and I've tried multiplying the RHS by its reciprocal and the LHS by cosA/(1+sinA). In that last attempt, I achieved an answer of 1=1, but I have a feeling that is incorrect and I'll need at least one trigonometric function in the solution.

If anyone wants to see the work that I've done in each attempt, just tell me and I'll post it.

Thanks for the assistance.

 

[tiny-tim]

Hi BlackHole213!

Either use the https://www.physicsforums.com/library.php?do=view_item&itemid=18" for A/2, or just multiply both sides by (1 - sinA + cosA).

 

[whyyie]

Always remember to multiply by the conjugate.

[PLAIN]http://img337.imageshack.us/img337/1976/50453339.jpg

 

[BlackHole213]

Thank you, both of you.