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

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Answer 1 · 2018-06-08T00:48:00

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

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 · 2018-06-08T00:55:28

See proof.

#(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#