NationStates Jolt Archive

Well, I'm probably going to fail the trig portion of my pre-cal exam, but I'd like to know how to solve this problem:

(sec^2(x) - 6tanx + 7)/(sec^2(x) - 5) = (tanx - 4)/(tanx + 2)

I've tried working with both sides, but I'm stumped no matter which one I try. Can anyone figure it out?

Well, I'm probably going to fail the trig portion of my pre-cal exam, but I'd like to know how to solve this problem:
(sec^2(x) - 6tanx + 7)/(sec^2(x) - 5) = (tanx - 4)/(tanx + 2)
I've tried working with both sides, but I'm stumped no matter which one I try. Can anyone figure it out?

hey Pie. well, have you ever considered working with a tutor? and do you ever look in the back of your textbook for the 'example' answers.. then try to figure them out, etc.? i wish you luck.. but there's no pie coming from this quarter..

Vegetarianistica']hey Pie. well, have you ever considered working with a tutor? and do you ever look in the back of your textbook for the 'example' answers.. then try to figure them out, etc.? i wish you luck.. but there's no pie coming from this quarter..

Meh, the rest of the exam I'll do fine on. And actually, I'm doing well enough with the other identities. THAT one just has me stumped beyond all reason. It's NOTHING like the other ones in the same question.

PS:

http://www.google.com/search?hl=en&lr=&q=online+trig+help&btnG=Search

sec ^(2) is 1+tan^(2)
or 1-tan^(2)
or tan^(2)-1, I am so bad at math.

It's a neat trick..

(sec^2(x) - 6tanx + 7)/(sec^2(x) - 5)
{-- sec(z) = 1/ cos(z) --}
= (1/cos(x)^2 - 6*sin(x)/cos(x) + 7)/(1/cos(x)^2 - 5)
{-- multiply both sides by cos^2(x) --}
= (1 - 6*sin(x)cos(x) + 7cos(x)^2)/(1 - 5cos(x)^2)
{-- 1=cos^2(x)+sin^2(x) --}
= (cos(x)^2+sin(x)^2 - 6*sin(x)cos(x) + 7cos(x)^2)/(1 - 5cos(x)^2)
{-- put terms together --}
= (sin(x)^2 - 6*sin(x)cos(x) + 8cos(x)^2)/(1 - 5cos(x)^2)
{-- factor --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/(1 - 5cos(x)^2)
{-- 1=cos^2(x)+sin^2(x) --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/(cos(x)^2+sin(x)^2 - 5cos(x)^2)
{-- put terms together --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/(sin(x)^2 - 4cos(x)^2)
{-- factor --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/[(sin(x)- 2*cos(x))(sin(x)+ 2*cos(x))]
{-- divide common factors out of expression --}
= (sin(x) - 4cos(x))/(sin(x) + 2cos(x) )
{-- divide both sides by cos(x) --}
= (sin(x)/cos(x) - 4)/(sin(x)/cos(x) + 2)
{-- tan(x) = sin(x)/cos(x) --}
= (tanx - 4)/(tanx + 2)

QED

One could still make it a bit neater of course.. (but two edits is my limit for today)

the answer: 42

Well, I'm probably going to fail the trig portion of my pre-cal exam, but I'd like to know how to solve this problem:

(sec^2(x) - 6tanx + 7)/(sec^2(x) - 5) = (tanx - 4)/(tanx + 2)

I've tried working with both sides, but I'm stumped no matter which one I try. Can anyone figure it out?

You use (x)? I'm stuck with Theta.

It's a neat trick..

(sec^2(x) - 6tanx + 7)/(sec^2(x) - 5)
{-- sec(z) = 1/ cos(z) --}
= (1/cos(x)^2 - 6*sin(x)/cos(x) + 7)/(1/cos(x)^2 - 5)
{-- multiply both sides by cos^2(x) --}
= (1 - 6*sin(x)cos(x) + 7cos(x)^2)/(1 - 5cos(x)^2)
{-- 1=cos^2(x)+sin^2(x) --}
= (cos(x)^2+sin(x)^2 - 6*sin(x)cos(x) + 7cos(x)^2)/(1 - 5cos(x)^2)
{-- put terms together --}
= (sin(x)^2 - 6*sin(x)cos(x) + 8cos(x)^2)/(1 - 5cos(x)^2)
{-- factor --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/(1 - 5cos(x)^2)
{-- 1=cos^2(x)+sin^2(x) --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/(cos(x)^2+sin(x)^2 - 5cos(x)^2)
{-- put terms together --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/(sin(x)^2 - 4cos(x)^2)
{-- factor --}
= (sin(x)- 2*cos(x))(sin(x)- 4*cos(x))/[(sin(x)- 2*cos(x))(sin(x)+ 2*cos(x))]
{-- divide common factors out of expression --}
= (sin(x) - 4cos(x))/(sin(x) + 2cos(x) )
{-- divide both sides by cos(x) --}
= (sin(x)/cos(x) - 4)/(sin(x)/cos(x) + 2)
{-- tan(x) = sin(x)/cos(x) --}
= (tanx - 4)/(tanx + 2)

QED

One could still make it a bit neater of course.. (but two edits is my limit for today)

Wow. Substituting (cos^2x + sin^2x) in for 1 is something I would have never thought of, even though it seems obvious now. Thanks!

You use (x)? I'm stuck with Theta.

Eh, it was a Theta, but I don't know how to make that symbol. x works well enough, a variable is a variable.

Wow. Substituting (cos^2x + sin^2x) in for 1 is something I would have never thought of, even though it seems obvious now. Thanks!
Yeah, sin(x)^2+cos(x)^2=1 is one of the most useful of all the trig identities you'll ever see. It's constantly used, so memorizing it and all of its forms is vital for sucess, especially in calc.