# Prove that the left hand side equals the right hand side: `(sinxcosx)/(1+cosx)` = `(1-cosx)/(tanx)`

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(There are numerous possibilities for proofs) To prove:

`(sinxcosx)/(1+cos x) = (1-cos x)/tan x`

We will manipulate the LHS (left-hand side). As we have (1+cosx) as the denominator multiply by `(1-cosx)/(1-cos x). ` This will not change the essence of the expression in any way as it is the equivalent of 1. It also suits our purposes (see RHS)

LHS: `therefore (sinxcosx)/(1+cosx) times (1-cos x)/(1-cosx)`

= `((sinxcosx)(1-cos x))/((1+cos x)(1-cosx))`

Simplify this by expanding the brackets regarding the numerator and denominator. The denominator is now a difference of two squares:

= `(sinxcosx- sincos^2x)/(1-cos^2x)`

We are manipulating this to suit our purposes so now factorize using sin (numerator) and the identity `1-cos^2x=sin^2x` in the denominator:

= `((sinx)(cos x-cos ^2 x))/(sin^2 x)`

The sin (numerator) cross cancels with one from the denominator and we factorize the remaining term:

`therefore =(cosx(1-cosx))/(sin x)`

We can rewrite (for our purpses) as:

`cosx/sinx times (1-cosx)/1`

We know that `sinx/cosx = tanx` so it follows that `cosx/sinx=1/tanx`

`therefore = 1/tanx times (1-cosx)/1`

`= (1-cosx)/tanx`

`therefore (sinxcosx)/(1+cosx)= (1-cosx)/tanx`

**Therefore LHS=RHS**