Question

prove the identity: (cosx/1-sinx)+(cosx/1+sinx) =2secx

Answer 1

find the common denominator

Answer 2

1-sin^2(x) ( DIFFERENCE OF SQUARES ) which turns in cos^2(x)

Answer 3

LHS should be 2cos(x)/cos^2(x)

Answer 4

first combine the fractions by getting a common denominator and adding

\[(cosx(1+sinx) + cosx(1-sinx))\div (1-sinx)(1+sinx)\]

now factor out the Cosx from the top and expand the bottom to get:

\[(cosx (1+ sinx + 1 - sinx)) \div (1 - Sin^2x)\]

simplifying we get:

\[cosx (2) \div (1-\sin^2x)\]

Using the Pathagorean Identity, we know that \[1-\sin^2 x = \cos^2x\]

so we can rewrite it as

\[(2)cosx \div \cos^2x\]

the top and bottom cancel to become

(2)/cosx = 2x (1/cosx)

since secx = 1/cosx

it becomes

2secx

Answer 5

sorry the last part shoudl say:

(2)/cosx = 2(1/cosx)

Answer 6

Hermeezey you have to give the student hints not solutions ...

Answer 7

hmm ok i just got stuck on the top part

Answer 8

which part do you mean?

Answer 9

what happend to the sinx's for the numerator?

Answer 10

Lol I see, i kinda skimmed that part i guess.

because (1+ sinx + 1 - sinx) can be reorganized as

(1 + 1 + sinx - sinx) and we have a +sinx and -sinx, they just cancel out, so you are left with a (1+1).

Answer 11

oh i see. ok i also so dont get how 2cosx became 2/cosx...

Answer 12

ok i understand now! thanks