Prove 1-sinA/1-secA - 1+sinA/1+secA = 2cotA(cosA-cosecA)

[hobomoe]

Homework Statement

Prove that
(1-sinA)/(1-secA) - (1+sinA)/(1+secA) = 2cotA(cosA-cosecA)

Homework Equations

tan^2+1=sec^2

The Attempt at a Solution

Times (1-sinA)/(1-secA) - (1+sinA)/(1-secA) = 2cotA(cosA-cosecA) by (1-sec^2 A) to get
(1-sinA)(1+secA) - (1+sinA)(1-secA) = 2cotA(cosA-cosecA)(tan^2 A)
= 2(sinA-secA)
(1-sinA)(1+secA) - (1+sinA)(1-secA) = (1+secA-sinA-sinAsecA)-(1-secA+sinA-sinAsecA)
= 2(secA-sinA)
I made at least one mistake in one of the steps I believe.

 

[sutupidmath]

Use more parentheses...

 

[hobomoe]

Better?

 

[sentient 6]

why are you assuming the two sides of the identity are equal in the proof itself?

 

[hobomoe]

They do equal. We're only asked to prove that they do, not to find out if they do or not.

 

[SammyS]

2cotA(cosA-cosecA)(tan^2 A)

(The right hand side of the second line of your solution) will be simpler if you recognize that cot(A)=1/tan(A).

Then use: tan(A) = sin(A)/cos(A).

Also: What sentient 6 points out in Post #4, is that when proving a trig identity, it is customary to not allow multiplying, dividing, adding, or subtracting both sides of the equation. Usually you are only allowed to work on each side independently. Very strict instructors will only allow you to work with one side or the other.

 

[hobomoe]

2cotA(cosA-cosecA)(tan^2 A)

(The right hand side of the second line of your solution) will be simpler if you recognize that cot(A)=1/tan(A).

Then use: tan(A) = sin(A)/cos(A).

I did do that to get 2cotA(cosA-cosecA)(tan^2 A) = 2(sinA-secA).
One tanA cancels out cotA and the other changes the cosA and cosecA.
Or am I wrong?

 

[SammyS]

I did do that to get 2cotA(cosA-cosecA)(tan^2 A) = 2(sinA-secA).
One tanA cancels out cotA and the other changes the cosA and cosecA.
Or am I wrong?

No, that's fine. I missed seeing that! DUH! (@ me)

What you have done looks fine. (except for the unusual way of working with identities)

Are you sure you didn't copy the problem wrong?

One way to check to see if an identity is true or false is to use a graphing calculator or program.

Graph the left side minus the right side. Graph should be y = 0 or very near zero, like y = 1×10^-15.

 

[hobomoe]

The answers are opposites of each other so where did I go wrong in one of the equations?
(ones positive and ones negative)

 

[SammyS]

I see the problem!

sec²(A) - 1 = tan(A)

Therefore: 1 - sec²(A) = -tan(A)

 

[hobomoe]

Oh haha big derp by me.. THANK YOU! Been stressing me out..