Question

How do you prove `(1+sec(x))/(sin(x)+tan(x))= csc(x)`?

Answer 1

I shall prove by using axioms and identities to change only one side of the equation until it is identical to the other side.

Explanation:

Given:

`(1+sec(x))/(sin(x)+tan(x))= csc(x)`

Substitute `tan(x) = sin(x)/cos(x)`:

`(1+sec(x))/(sin(x)+sin(x)/cos(x))= csc(x)`

Substitute `sec(x) = 1/cos(x)`:

`(1+1/cos(x))/(sin(x)+sin(x)/cos(x))= csc(x)`

Factor the denominator:

`(1+1/cos(x))/(sin(x)(1+1/cos(x)))= csc(x)`

Please observe that there is a common factor in the numerator and denominator that cancels:

`1/sin(x)= csc(x)`

Use the identity `1/sin(x) = csc(x)`:

`csc(x)= csc(x)` Q.E.D.

Answer 2

See proof.

Explanation:

`(1+secx)/(sinx+tanx)= cscx`

`(1+1/cos)/(sinx+sinx/cosx)= cscx`

`(cosx/cosx+1/cosx)/((sinxcosx)/cosx+sinx/cosx)= cscx`

`((cosx+1)/cosx)/((sinxcosx+sinx)/cosx)= cscx`

`(cosx+1)/cosx*cosx/(sinxcosx+sinx)= cscx`

`(cosx+1)/cancel(cosx)*cancel(cosx)/(sinxcosx+sinx)= cscx`

`(cosx+1)/(sinxcosx+sinx)= cscx`

`(cosx+1)/(sinx(cosx+1))= cscx`

`cancel(cosx+1)/(sinxcancel(cosx+1))= cscx`

`1/sinx=cscx`

`cscx=cscx`

`:. (1+secx)/(sinx+tanx)= cscx`