Question

I have never been able to solve a trig identity before. sec^2x-2secx cosx +cos^2x = tan^2x - sin^2x

Answer 1

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

Answer 2

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).

Answer 3

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)

Answer 4

... Is that your solution?

Answer 5

no...just trying to motivate you :)

next, i would convert everything to sin x and cos x

Answer 6

right now, i'd have:
(sec x - cos x)^2 = (tan x - sin x)(tan x + sin x)

Answer 7

I don't know anything to be motivated lol.
And I don't understand why your using both sides of the equation

Answer 8

@jim_thompson5910, wanna give us a hand?

Answer 9

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

Answer 10

(1/cos - cos)^2 = (tan - sin)(tan + sin)

Answer 11

(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)

Answer 12

... what now?

Answer 13

sin^4/cos^2 = tan^2 - sin^2

Answer 14

sin^2*sin^2/cos^2 = tan^2 - sin^2

Answer 15

sin^2(1-cos^2)/cos^2 = tan^2 - sin^2

Answer 16

(sin^2 - sin^2cos^2)/cos^2 = tan^2 - sin^2

Answer 17

sin^2/cos^2 - sin^2cos^2/cos^2 = tan^2 - sin^2

Answer 18

tan^2 - sin^2 = tan^2 - sin^2

Answer 19

@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)?…

Answer 20

as required...so what i did on the right initially was unnecessary...but i didn't know that until the end

Answer 21

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

Answer 22

\[\large \sec^2(x) - 2\sec(x)\cos(x) + \cos^2(x) = \tan^2(x) - \sin^2(x)\]


\[\large \sec^2(x) - 2*\frac{1}{\cos(x)}\cos(x) + \cos^2(x) = \tan^2(x) - \sin^2(x)\]


\[\large \sec^2(x) - 2*\frac{\cos(x)}{\cos(x)} + \cos^2(x) = \tan^2(x) - \sin^2(x)\]


\[\large \sec^2(x) - 2 + \cos^2(x) = \tan^2(x) - \sin^2(x)\]


\[\large 1+\tan^2(x) - 2 + 1 - \sin^2(x) = \tan^2(x) - \sin^2(x)\]


\[\large \tan^2(x) - \sin^2(x) = \tan^2(x) - \sin^2(x)\]


The identity is confirmed.

Answer 23

Okay, if I could give a billion medals to both of you right now, I would.
ahhhh! thank you guys sfm D:

Answer 24

glad to be of help

Answer 25

hopefully i didn't confuse you too much...i was working it on the fly. :)