How do you prove #sec x + sin x = tan x sin x#?

1 Answer
Apr 26, 2016

This is true only for #x=-\pi/4# plus multiples of #\pi#.

Explanation:

Since #\sec(x)=1/\cos(x)# and #\tan(x)=\sin(x)/\cos(x)#, first multiply by #\cos(x)#. Thus

#1+\sin(x)\cos(x)=\sin^2(x)#

Put in #1=\sin^2(x)+\cos^2(x):

#\sin^2(x)+\cos^2(x)+\sin(x)\cos(x)=\sin^2(x)#

#\cos^2(x)+\sin(x)\cos(x)=0#

#\cos(x)(\cos(x)+\sin(x))=0#

Thus the equation is true only if #\cos(x)=0# or #\cos(x)=-sin(x)#.

But we cannot accept #\cos(x)=0# because that leaves #\sec(x)=1/\cos(x)# undefined. So #\cos(x)=-sin(x)#, thus #x=-\pi/4# plus multiples of #\pi#.