Prove that #(cosxcotx)/(1 - sinx) - 1 = cscx#?
2 Answers
We start with:
#color(green)((cosxcotx)/(1-sinx) - 1 stackrel(?)(=) cscx).#
First, recall
#(cosx*cosx/sinx)/(1-sinx) - 1 = cscx#
#(cos^2x/sinx)/(1-sinx) - 1 = cscx#
On the fraction you can move the middle
#cos^2x/(sinx(1-sinx)) - 1 = cscx#
Now, when you have a
#cos^2x/(sinx(1-sinx)) - (sinx(1-sinx))/(sinx(1-sinx)) = cscx#
Distribute the numerator and combine into one fraction:
#(-sinx + cos^2x + sin^2x)/(sinx(1-sinx)) = cscx#
Then recall
#cancel(1 - sinx)/(sinxcancel((1-sinx))) = cscx#
#1/sinx = cscx#
#color(blue)(cscx = cscx)#
It is a bit long but...
Explanation:
You can try changing all in
Your identity becomes:
we use
and:
but:
so we get: