Two trig Identites:  a) cosx/1-sinx - cosx/1+sinx =2 tanx b) 1-sinx/cosx = sinx/1+cosxMost of these identities hold true but some of them are false. Prove LS = RS or show why the rule does not hold

a) You need to bring the terms to a common denominator such that:

`(cos x*(1+sin x) - cos x*(1 - sin x))/((1+sin x)(1 - sin x)) = 2 tan x`

You need to open the brackets such that:

`(cos x + cos x*sin x - cos x + cos x* sin x)/(1 - sin^2 x) = 2 tan x`

Reducing like terms yields:

`(2cos x*sin x)/(1 - sin^2 x) = 2 tan x`

You need to remember that `1 - sin^2 x = cos^2 x`  such that:

`(2cos x*sin x)/(cos^2 x) = 2 tan x`

Reducing by `cos x ` yields:

`(2sin x)/(cos x) = 2 tan x`

You need to substitute `sin x/cos x`  by tan x such that:

`2 tan x = 2 tan x`

Hence, using formulas of trigonometry and the laws applied to difference of fractions that have different denominators yields that `cosx/(1-sinx) - cosx/(1+sinx) =2 tanx` .

b) You need to multiply by `cos x(1 + cos x)`  both sides such that:

`(1 - sin x)(1 + cos x) = sin x*cos x`

You need to open the brackets such that:

`1 + cos x - sin x - sin x*cos x = sin x*cos x`

`1 + cos x - sin x = 2 sin x*cos x`

`1 + cos x - sin x = sin 2x`

Hence, making transformations to the left side yields that the expression `(1-sinx)/cosx != sinx/(1+cosx).`