#(sec(x) + csc(x))/(1+ tan(x))# express in one trigonometric function?

1 Answer
Nov 20, 2017

#(sec(x) + csc(x))/(1+ tan(x)) = csc(x)#

Explanation:

Given:

#(sec(x) + csc(x))/(1+ tan(x)) = #

Substitute #1/cos(x)# for #sec(x)#, #1/sin(x)# for #csc(x)#, and #sin(x)/cos(x)# for #tan(x)#:

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

Make common denominators:

#(sin(x)/(sin(x)cos(x)) + cos(x)/(sin(x)cos(x)))/(cos(x)/cos(x)+ sin(x)/cos(x)) = #

Combine the numerators over the common denominators:

#((sin(x) + cos(x))/(sin(x)cos(x)))/((cos(x) + sin(x))/cos(x)) = #

Division of fractions is the same as multiplication by the reciprocal of the divisor:

#(sin(x) + cos(x))/(sin(x)cos(x))(cos(x))/(cos(x) + sin(x)) = #

Please observe the factors that cancel:

#(cancel(sin(x) + cos(x)))/(sin(x)cancel(cos(x)))(cancel(cos(x)))/(cancel(cos(x) + sin(x))) = #

Remove the cancelled factors:

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