How do you prove $2tan(x)sec(x) = (1/(1-sin(x))) - (1/(1+sin(x)))$ ?

Question

14996 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-26T08:20:36

Please see explanation.

Here,
$2tanxsecx=1/(1-sinx)-1/(1+sinx)$
We take,

$RHS=1/(1-sinx)-1/(1+sinx)$

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

$=(cancel1+sinx-cancel1+sinx)/(1-sin^2x)$

$=(2sinx)/cos^2x$

$=2sinx/cosx1/cosx$ , where, $sinx/cosx=tanx and 1/cosx=secx$

$=2tanxsecx$ ,

$=LHS$

How do you prove 2tan(x)sec(x) = (1/(1-sin(x))) - (1/(1+sin(x)))2tan(x)sec(x) = (1/(1-sin(x))) - (1/(1+sin(x)))?