How do you simplify #(1 - (sinx)^2) / (sinx - cscx)#?
1 Answer
Jun 15, 2016
-sinx
Explanation:
Let's begin by simplifying the denominator of the fraction.
using
#cscx=1/(sinx)" we obtain"#
#sinx-1/sinx#
which we require to express as a single fraction.
#sinx xx(sinx/sinx)-1/sinx=sin^2x/sinx-1/sinx# We now have a common denominator of sinx
#rArrsin^2x/sinx-1/sinx=(sin^2x-1)/sinx=(-(1-sin^2x))/sinx# The overall fraction is now
#(1-sin^2x)/((-(1-sin^2x))/(sinx))# We can now invert the denominator and multiply
#cancel(1-sin^2x)^1 xx(sinx)/-cancel((1-sin^2x)^1)#
#=sinx/(-1)=-sinx#
