How do you prove #sin(x)(cos(x))/(cos^2(x) - sin^2(x)) = tan(x)/(1 - tan(x))#?

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Answer 1 · 2016-02-13T03:55:39

The identity is false.

The given identity is not true. For example, if #x = pi/6#

#(sin(pi/6)cos(pi/6))/(cos^2(pi/6)-sin^2(pi/6)) = (1/2*sqrt(3)/2)/(3/4-1/4)=(sqrt(3)/4)/(2/4)=sqrt(3)/2#

but

#tan(pi/6)/(1-tan(pi/6)) = (1/sqrt(3))/(1-1/sqrt(3))=1/(sqrt(3)-1) != sqrt(3)/2#

In fact, the equation is only true when #sin(x) = 0# as can be seen here:

#tan(x)/(1-tan(x)) = (sin(x)/cos(x))/(cos(x)/cos(x)-sin(x)/cos(x))#

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

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

#=(sin(x)cos(x) + sin^2(x))/(cos^2(x)-sin^2(x))#

#=(sin(x)cos(x))/(cos^2(x)-sin^2(x))+sin^2(x)/(cos^2(x)-sin^2(x))#

thus the equation holds only when #sin^2(x)/(cos^2(x)-sin^2(x))=0#
which is true only when #sin(x) = 0#