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

2 Answers
Douglas K.
Jun 8, 2018

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

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

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

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

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

Factor the denominator:

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

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

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

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

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

Bill Jorgensen
Jun 8, 2018

See proof.

Explanation:

(1+secx)/(sinx+tanx)= cscx1+secxsinx+tanx=cscx

(1+1/cos)/(sinx+sinx/cosx)= cscx1+1cossinx+sinxcosx=cscx

(cosx/cosx+1/cosx)/((sinxcosx)/cosx+sinx/cosx)= cscxcosxcosx+1cosxsinxcosxcosx+sinxcosx=cscx

((cosx+1)/cosx)/((sinxcosx+sinx)/cosx)= cscxcosx+1cosxsinxcosx+sinxcosx=cscx

(cosx+1)/cosx*cosx/(sinxcosx+sinx)= cscxcosx+1cosx⋅cosxsinxcosx+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

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