# Prove the identity: (cosx/1-sinx) - (cosx/1+sinx) = 2tanx ------------------------------- Thank you!

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

First, move cosx from the brackets:

cosx/(1-sinx) - cosx/(1+sinx)=

cosx*(1/(1-sinx) - 1/(1+sinx)).

Then find and use the common denominator:

= cosx*(1+sinx - (1-sinx))/((1-sinx)*(1+sinx)).

Take into account that 1-(sinx)^2={cosx)^2 obtain

= cosx*2*sinx/(cosx)^2 = 2*sinx/cosx

which is = 2*tanx by the definition of tanx.

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kspcr111 | In Training Educator

Posted on

To prove

`(cosx/(1-sinx)) - (cosx/(1+sinx)) = 2*tanx`

Sol:

`(cosx/(1-sinx)) - (cosx/(1+sinx)) = ((cosx)*(1+sinx) -(cosx)*(1-sinx))/((1+sinx)*(1-sinx))`

= `(2*cosx*sinx)/((1+sinx)*(1-sinx))`

`=(2*cosx*sinx)/(1-sin^2x)`

`=(2*cosx*sinx)/(cos^2x)                                                 =(2*sinx)/(cosx)                                                 =2*tanx`

dnelson0107 | High School Teacher | eNotes Newbie

Posted on

Multiply both expression by the other denominator to achieve a common denominator.

cosx(1+sinx)/(1-sinx)(1+sinx) - cosx(1-sinx)/(1+sinx)(1-sinx) =

coxx(1+sinx)/(1-sin^2x) - cosx(1-six)/(1-sin^2x)

Simplify by Pythagorean Identity (1-sin^2x) = cos^2x

cosx(1+sinx)/cos^2x - cosx(1-sinx)/cos^2x

Simplify cancelling out a cosx from each term

(1+sinx)/cosx - 1-sinx/cosx

Now combine fractions

[1+sinx - (1 - sinx)]/cosx

= (1 + sinx - 1 + sinx)/cosx

= 2sinx/cosx = 2tanx