# Show that (1 + sin x)/(1-sin x)-(1 - sin x)/(1 + sin x) = 4 tan x sec x

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The trigonometric identity `(1 + sin x)/(1-sin x)-(1 - sin x)/(1 + sin x) = 4 tan x sec x ` has to be proved.

Start with the left hand side:

`(1 + sin x)/(1-sin x)-(1 - sin x)/(1 + sin x)`

Multiply to get a common denominator

= `((1 + sin x)(1 + sin x) - (1 - sin x)(1 - sin x))/((1-sin x)(1 + sin x))`

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

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

= `(4*sin x)/((1-sin x)(1 + sin x))`

= `(4*sin x)/(1-sin^2x)`

Use the identity `sin^2x +cos^2x = 1` or `cos^2x = 1 - sin^2x`

= `(4*sin x)/(cos^2x)`

= `4*(sin x/cos x)*(1/cos x)`

= `4*tan x*sec x`

**This proves that **`(1 + sin x)/(1-sin x)-(1 - sin x)/(1 + sin x) = 4 tan x sec x`

To prove the relation is the same as proving the identity:

`(1 + sin x)/(1-sin x)-(1 - sin x)/(1 + sin x) = 4 tan x sec x`

Now, we can first rewrite the RHS as (using tan x = (sinx)/(cosx) and sec x = 1/cosx):

`4*(sin x)/(cos^2x)`

On the other hand, the LHS is (putting all with common denominator):

`((1+sin x)^2-(1-sin x)^2)/(1-sin^2 x)=4*(sin x)/(1-sin^2 x)=4*(sin x)/(cos^2 x)` . QED