Question

Prove that 1+sinx/1-sinx -1-sinx/1+sinx=4tanxsecx

Answer 1

If interpreted as shown, the order is shown in the lines following
1+sin(x)/1-sin(x)-1-sin(x)/1+sin(x)
=1 + ((sin(x))/1)  - sin(x) -1 - (sin(x)/1)  + sin(x)
which is completely meaningless for the purpose of this question.

The left-hand side is
$\frac{1+sin(x)}{1-sin(x)}-\frac{1-sin(x)}{1+sin(x)}$
common denominator is (1-sin(x))(1+sin(x)), so
$=\frac{(1+sin(x))(1+sin(x)-(1-sin(x))(1-sin(x))}{(1+sin(x)(1-sin(x)}$
$=\frac{(1+sin(x))^2-(1-sin(x))^2}{(1+sin(x)(1-sin(x)}$
$=\frac{(1+1)(2sin(x))}{(1-sin^2(x)}$  using a^2-b^2=(a+b)(a-b)
$=\frac{4sin(x)}{cos^2(x)}$
$=\frac{4tan(x)}{cos(x)}$
$=4tan(x)sec(x)$

Therefore 
$\frac{1+sin(x)}{1-sin(x)}-\frac{1-sin(x)}{1+sin(x)}=4tan(x)sec(x)$