Question

proving identities:
(cot+cos)/(sec-cos) = (csc + 1) / tan^2

Answer 1

(cot x+cos x)/(sec x -cos x) = (csc x + 1) / tan^2

( [cos x / sin x] + cos x ) / ( [1 /cos x] - cos x ) = (csc x + 1) / tan^2

( [cos x / sin x] + cos x ) / ( [1 /cos x] - cos^2x /cos x ) = (csc x + 1) / tan^2

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^
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so far so good?

Answer 2

wait. i'm lost. i dont get it
@SolomonZelman

Answer 3

Alright, do you get the second line that I wrote?

Answer 4

(cot x+cos x)/(sec x -cos x) = (csc x + 1) / tan^2 <--- The initial problem.





Translating the left side into cosines and sines ---->

( [cos x / sin x] + cos x ) / ( [1 /cos x] - cos x ) = (csc x + 1) / tan^2


Are you good with this so far?

Answer 5

i dont get the 2nd line especially the part where the arrow is pointed at

Answer 6

So my first step was translating the left hand side into sines and cosines, where I got the following equation below.

( [cos x / sin x] + cos x ) / ( [1 /cos x] - cos x ) = (csc x + 1) / tan^2

Answer 7

then the arrow pointing is just to attract your attention to where am I changing the equation,
I am making cos x into cos^2x/cosx to add inside the second parenthesis.

Answer 8

okay got it now @SolomonZelman
then what?