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

2 Answers
Y
May 10, 2017

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)

Monzur R.
May 10, 2017

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

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