anonymous
  • anonymous
I have never been able to solve a trig identity before. sec^2x-2secx cosx +cos^2x = tan^2x - sin^2x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
the first thing i see is that the part on the left looks a lot like (a - b)^2 = a^2 -2ab + b^2 so you can factor the left into (sec x - cos x)^2
anonymous
  • anonymous
can u please help me here??? what of the circle that has a diameter endpoints are (2,7) and -6-1) (4,3), (-2,4), or (-2,3)? 2nd question. finde the distance between points (-4,-5) and (3,-1).
anonymous
  • anonymous
also, the right side, looks like a^2 - b^2 which factors as a difference of squares so the right factors to (tan x - sin x)(tan x + sin x)

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anonymous
  • anonymous
... Is that your solution?
anonymous
  • anonymous
no...just trying to motivate you :) next, i would convert everything to sin x and cos x
anonymous
  • anonymous
right now, i'd have: (sec x - cos x)^2 = (tan x - sin x)(tan x + sin x)
anonymous
  • anonymous
I don't know anything to be motivated lol. And I don't understand why your using both sides of the equation
anonymous
  • anonymous
@jim_thompson5910, wanna give us a hand?
anonymous
  • anonymous
i'm using both sides so that i can get to things that lok better...sin and cos are easier to work with, IMO, than sec and tan
anonymous
  • anonymous
(1/cos - cos)^2 = (tan - sin)(tan + sin)
anonymous
  • anonymous
(1/cos - cos^2/cos)^2 = (tan - sin)(tan+sin) ((1-cos^2)/cos)^2 = (tan - sin)(tan + sin) (sin^2/cos)^2 = (tan - sin)(tan + sin)
anonymous
  • anonymous
... what now?
anonymous
  • anonymous
sin^4/cos^2 = tan^2 - sin^2
anonymous
  • anonymous
sin^2*sin^2/cos^2 = tan^2 - sin^2
anonymous
  • anonymous
sin^2(1-cos^2)/cos^2 = tan^2 - sin^2
anonymous
  • anonymous
(sin^2 - sin^2cos^2)/cos^2 = tan^2 - sin^2
anonymous
  • anonymous
sin^2/cos^2 - sin^2cos^2/cos^2 = tan^2 - sin^2
anonymous
  • anonymous
tan^2 - sin^2 = tan^2 - sin^2
anonymous
  • anonymous
@mtbender74 can u please help me here? what of the circle that has a diameter endpoints are (2,7) and -6-1) (4,3), (-2,4), or (-2,3)?…
anonymous
  • anonymous
as required...so what i did on the right initially was unnecessary...but i didn't know that until the end
jim_thompson5910
  • jim_thompson5910
The identities I'm going to use are \[\large \sec(x) = \frac{1}{\cos(x)}\] \[\large \sec^2(x) = 1+\tan^2(x)\] \[\large \cos^2(x) = 1-\sin^2(x)\] These identities (and many more) can be found here http://www.sosmath.com/trig/Trig5/trig5/trig5.html
jim_thompson5910
  • jim_thompson5910
\[\large \sec^2(x) - 2\sec(x)\cos(x) + \cos^2(x) = \tan^2(x) - \sin^2(x)\] \[\large \sec^2(x) - 2*\frac{1}{\cos(x)}\cos(x) + \cos^2(x) = \tan^2(x) - \sin^2(x)\] \[\large \sec^2(x) - 2*\frac{\cos(x)}{\cos(x)} + \cos^2(x) = \tan^2(x) - \sin^2(x)\] \[\large \sec^2(x) - 2 + \cos^2(x) = \tan^2(x) - \sin^2(x)\] \[\large 1+\tan^2(x) - 2 + 1 - \sin^2(x) = \tan^2(x) - \sin^2(x)\] \[\large \tan^2(x) - \sin^2(x) = \tan^2(x) - \sin^2(x)\] The identity is confirmed.
anonymous
  • anonymous
Okay, if I could give a billion medals to both of you right now, I would. ahhhh! thank you guys sfm D:
jim_thompson5910
  • jim_thompson5910
glad to be of help
anonymous
  • anonymous
hopefully i didn't confuse you too much...i was working it on the fly. :)

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