Question

How do I prove that `(tanx-sinx)/(tanxsinx)=(1-cosx)/sinx`?

Answer 1

See below.

Explanation:

Formatted question: Prove `(1-cosx)/sinx=(tanx-sinx)/(tanxsinx)`

Since the right-hand side (RHS) appears more complicated, we will start with that side.

Consider that `tanx=sinx/cosx`. We can substitute this in for the `tanx` in the numerator and denominator:
`RHS=((sinx/cosx)-sinx)/((sinx/cosx)sinx)`

By simplifying:
`=((sinx/cosx)-(sinxcosx)/cosx)/(sin^2x/cosx)`
`=((sinx-sinxcosx)/cosx)/(sin^2x/cosx)`
`=(((sinx(1-cosx))/cosx)*(cosx/sin^2x)`

We can cancel one of `sinx`'s and the `cosx` from the numerator and denominator since we are multiplying the two fractions:
`=(1-cosx)/sinx=LHS`

`therefore` `(1-cosx)/sinx=(tanx-sinx)/(tanxsinx)`

Answer 2

See below

Explanation:

Since proofs work both ways, we can start from the `sf(RHS)` and prove the `sf(LHS)`.

`(tan x-sinx)/(tan x sinx)`

`=tanx/(tan x sinx)-sinx/(tan x sinx)`

`=1/sinx-1/tanx`

`=1/sinx-cosx/sinx`

`=(1-cosx)/sinx`

`therefore(tanx-sinx)/(tanxsinx)-=(1-cosx)/sinx`, as required. `square`