Prove the identity (cosx)/(1=sinx) + (cosx)/(1-sinx)=2secx?

Question

Step by step please

its (cosx)/(1+sinx) not 1=sinx

Can you write out what your doing, I dont see whats going on and both answers are difficult to follow

Answer 1

cosx/(1+sinx)+cosx/(1-sinx)
cosx(1-sinx)+cosx(1+sinx)/
(1+sinx)(1-sinx)
=cosx-cosxsinx+cosx+cosxsinx/
(1-sin^2x)
=2cosx/cos^2x
=2/cosx
=2secx

Answer 2

LHS:
(cosx)/(1+sinx) + (cosx)/(1-sinx)
=(cos x -sin x cos x + sin x cos x + cos x)/(1-sin^2 x)
=2cos x/cos^2 x
=2/cos x
=2secx

Answer 3

sin(x) / cos(x) = tan(x) a million + tan(x) + cos(x)/(a million + sin(x) ) = 2 sec(x) a million + tan(x) + cos(x) [a million - sin(x) / (a million - sin^2(x) ) = 2 sec(x) a million + tan(x) + cos(x) [a million - sin(x) ) / cos^2(x) = 2 sec(x) a million + tan(x) + (a million - sin(x) ) /cos(x) a million + tan(x) + sec(x) - tan(x) a million + sec(x) And it does no longer. they do now no longer seem to be equivalent.

Answer 4

((cos(x))/(1 + sin(x))) + ((cos(x))/(1 - sin(x))) = 2sec(x)

Multiply everthing on the left side by (1 - sin(x))(1 + sin(x))

((cos(x))(1 - sin(x)) + (cos(x))(1 + sin(x)))/(1 - sin(x)^2) = 2sec(x)

((cos(x))((1 - sin(x)) + (1 + sin(x)))/(cos(x)^2) = 2sec(x)

(1 - sin(x) + 1 + sin(x))/(cos(x)) = 2sec(x)

2/(cos(x)) = 2sec(x)

this is the same as just saying

2sec(x)

For more info, go to

http://www.math.com/tables/trig/identities.htm