# straight up hard trig problem

#### eatkabab

##### Full Member
10+ Year Member
got this from a friend that couldn't solve it...

Sec^2 (A) + Tan^2 (A-3) = 0

apparently the answer is (pi/4) but I'm not 100% sure of that cuz it don't make any sense!!! I've been lookin at this problem and trying to figure it out for a month now. still can't get it...

I just attempted it again and got stuck here:

1 = tan^2 (A-(Pi/60)) - tan^2 (A)
1 = tan (A) [tan (1-(Pi/60A)) - tan (Pi/180)]

I don't know if anything I'm doing is correct by the way. I'm not even sure if I can convert that "A-3" into radians from the beginning...

along with this problem, I have a piece of it that might be a step along the way?

2sin (A-1) = 0

#### Streetwolf

##### Ultra Senior Member
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7+ Year Member
got this from a friend that couldn't solve it...

Sec^2 (A) + Tan^2 (A-3) = 0

I'm thinking of an identity here. It's tan^2(A) + 1 = sec^2(A).

So you get tan^2(A) + 1 + tan^2(A-3) = 0

And so tan^2(A) + tan^2(A-3) = -1

Now you see why you can't solve it. You are looking for two squares which add up to -1. Unless you are dealing with imaginary numbers (which you aren't), you aren't going to solve this problem.

There must be some error in the problem.

#### eatkabab

##### Full Member
10+ Year Member
swear thats exactly what I said in the first 5 min that I looked at the problem a month ago!!! I thought I was crazy or somethin. whew... thanx, at least I'm not alone

#### doc3232

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I'm thinking of an identity here. It's tan^2(A) + 1 = sec^2(A).

So you get tan^2(A) + 1 + tan^2(A-3) = 0

And so tan^2(A) + tan^2(A-3) = -1

Now you see why you can't solve it. You are looking for two squares which add up to -1. Unless you are dealing with imaginary numbers (which you aren't), you aren't going to solve this problem.

There must be some error in the problem.

The A-3 looks fishy if you know what I mean (mistype by the book, not eatk).
Anyone know if I need to memorize all the old trig identities?
I don't want to. #### eatkabab

##### Full Member
10+ Year Member
ya man, there are only a few trig identities. you should know them:

sin^2(X) + cos ^2(X) = 1

1 + cot^2(X) = csc^2(X)
1+ tan^2(X) = sec^2(X)

sin(2X) = 2sinXcosX

cos(2X) = cos^2(X) - sin^2(X)
cos(2X) = 1 - 2sin^2(X)
cos(2X) = 2cos^2(X) - 1

sec(Pi/4) = csc(Pi/4) = root(2)

and the rest is basic stuff...

lemme know if I missed any

#### allstardentist

##### All-Star
10+ Year Member
A-3 must be a typo. Also, you wont get any ridiculously hard trig problems on the real test. just know the basic trig properties and identities, then u should be fine.

#### americanpierg

##### Senior Member
10+ Year Member
5+ Year Member
I'm thinking of an identity here. It's tan^2(A) + 1 = sec^2(A).

So you get tan^2(A) + 1 + tan^2(A-3) = 0

And so tan^2(A) + tan^2(A-3) = -1

Now you see why you can't solve it. You are looking for two squares which add up to -1. Unless you are dealing with imaginary numbers (which you aren't), you aren't going to solve this problem.

There must be some error in the problem.

Um no. That -1 just means you used the wrong identity. Add it to the other side, change the tan^2(A) + 1 into sec^2(A) and go from there. The problem is tha that was actually the ORIGINAL problem, so you took an extra step that did nothing for you.

#### Streetwolf

##### Ultra Senior Member
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7+ Year Member
Um no. That -1 just means you used the wrong identity. Add it to the other side, change the tan^2(A) + 1 into sec^2(A) and go from there. The problem is tha that was actually the ORIGINAL problem, so you took an extra step that did nothing for you.
The original problem as he wrote it is Sec^2 (A) + Tan^2 (A-3) = 0. One trig identity is Tan^2(A) + 1 = Sec^2(A).

By subbing in Tan^2(A) + 1 into the original problem in place of Sec^2(A), you get Tan^2(A) + 1 + Tan^2(A-3) = 0 which becomes Tan^2(A) + Tan^2(A-3) = -1.

Now you tell me a way to square two numbers, add them together, and get a negative number.

#### Simiam

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2i^2 = -4 right? I think that is calculus though, which is beyond the scope of the DAT

i^2 = -1

#### Streetwolf

##### Ultra Senior Member
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Yes I stated in my first post that I'm not considering imaginary numbers.

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