Solve Identity: 1/(cosA+sinA) + 1/(cosA-sinA)=tan2AcosecA

[brandon26]

Please help me prove this identity:

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.
I can't go any further, please help!

 

[arildno]

Well, how is cos^2A-sin^2A related to cos2A?

 

[TD]

And then for the RHS, try changing the tan and the csc to the corresponding sine and/or cosine expressions.

 

[brandon26]

Well, how is cos^2A-sin^2A related to cos2A?[/QUOT
one side is equal to the other.

 

[brandon26]

I got as far as simplifying the equation to 2cosA / cos2A. What now?

 

[quantumdude]

Please help me prove this identity:

1/(cosA+sinA) + 1/(cosA-sinA) = tan2AcosecA

I got as far as LHS= 2cosA/cos^2A - sin^2A.

So far so good. To finish you'll need the following trig identity:

\tan(2A)=\frac{2\tan(A)}{1-\tan^2(A)}

You'll want to try to make the LHS of your identity look like that. To do that you'll need to divide the numerator and denominator of your expression by some other expression, and you need to figure out which one that is.

Hint: Look at the 1 in the denominator above. What would you have to divide \cos^2(A) by to get a 1?

 

[arildno]

The simplest way is now to see:
\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)

I'll leave the last step to you..

 

[TD]

I got as far as simplifying the equation to 2cosA / cos2A. What now?

Once you have this, substitute tan(2a) by sin(2a)/cos(2a) and use the double angle formula on sin(2a). After that, realize that csc(a) = 1/sin(a) and you should be there

 

[brandon26]

The simplest way is now to see:
\frac{2\cos(A)}{\cos(2A)}=1*\frac{2\cos(A)}{\cos(2A)}=\frac{\sin(2A)}{\sin(2A)}*\frac{2\cos(A)}{\cos(2A)}=\frac{2\cos(A)}{\sin(2A)}\tan(2A)
I'll leave the last step to you..

tan2A (2cosA/sin2A) = tan2A (2cosA/ (2sinAcosA)) = tan2AcosecA.

Safe boys!