How do you simplify #tan x cos x csc x#?

1 Answer
Jul 22, 2015

It equals 1 for all values of #x# where each factor is defined.

Explanation:

Using the definition of tangent and cosecant gives:

#tan(x)cos(x)csc(x)=(sin(x))/(cos(x))*cos(x)*1/(sin(x))#

Everything now cancels to give

#tan(x)cos(x)csc(x)=(cancel(sin(x)))/(cancel(cos(x)))*cancel(cos(x))*1/(cancel(sin(x)))=1#

for all values of #x# where each of the original factors is defined.

The values of #x# where this is not true are those values of #x# which make either #cos(x)=0# or #sin(x)=0#. One of these will happen at each value of #x# that is an integer multiple of #pi/2# radians (90 degrees).

Hence, #tan(x)cos(x)csc(x)=1# for all #x# except #x=(n pi)/2#, where #n=0,\pm 1, \pm 2, \pm 3,...#