# Double/half angle identity help!

• February 12th 2013, 10:02 AM
DonGorgon
Double/half angle identity help!
Hello all, currently working on an assignment proving identities via the usage of half-angle and double-angle formulas.
All have seemed ok so far, yet I'm incredibly stuck on this problem:

cosA+sinA/cosA-sinA=sec2A + tan2A

Any help would be greatly appreciated
• February 12th 2013, 11:01 AM
emakarov
Re: Double/half angle identity help!
Start by learning the order of operations and correct the identity to be proved. Then express sec and tan through sin and cos, use the formulas for sin(2A) and cos(2A), and multiply the numerator and the denominator of the left-hand side by cos(A) + sin(A).
• February 12th 2013, 08:50 PM
Soroban
Re: Double/half angle identity help!
Hello, DonGorgon!

Quote:

$\text{Prove: }\:\frac{\cos A+\sin A}{\cos A-\sin A} \:=\:\sec2A + \tan2A$

$\text{Multiply the left side by }\frac{\cos A + \sin A}{\cos A + \sin A}:$
. . $\frac{\cos A + \sin A}{\cos A - \sin A}\cdot\frac{\cos A + \sin A}{\cos A + \sin A} \:=\: \frac{\overbrace{\cos^2\!A+\sin^2\!A}^{\text{This is 1}} + \overbrace{2\sin A\cos A}^{\text{This is }\sin2A}}{\underbrace{\cos^2\!A -\sin^2\!A}_{\text{This is }\cos2A}}$

. . $=\;\frac{1 + \sin2A}{\cos2A} \;=\;\frac{1}{\cos2A} + \frac{\sin2A}{\cos2A} \;=\;\sec2A + \tan2A$