How do I prove that #cscx-sinx=cosxcotx#?
↳Redirected from
"What is a supersaturated solution?"
#cscx-sinx#
#=1/sinx-sinx#
#=(1-sin^2x)/sinx#
#=cos^2x/sinx#
#=cosx/sinx xxcosx#
#=cotxcosx#
#thereforecscx-sinx=cotxcosx# #sf(QED)#
We need
#cscx=1/sinx#
#sin^2x+cos^2x=1#
#cotx=cosx/sinx#
Therefore,
#LHS=cscx-sinx#
#=1/sinx-sinx#
#=(1-sin^2x)/sinx#
#=cos^2x/sinx#
#=cosx*cosx/sinx#
#=cosxcotx#
#=RHS#
#QED#
We have: #csc(x) - sin(x)#
Let's apply a standard trigonometric identity; #csc(x) = frac(1)(sin(x))#:
#= frac(1)(sin(x)) - sin(x)#
#= frac(1)(sin(x)) - frac(sin^(2)(x))(sin(x))#
#= frac(1 - sin^(2)(x))(sin(x))#
One of the Pythagorean identities is #cos^(2)(x) + sin^(2)(x) = 1#
We can rearrange it to get:
#Rightarrow cos^(2)(x) = 1 - sin^(2)(x)#
Let's apply this rearranged identity to our proof:
#= frac(cos^(2)(x))(sin(x))#
#= cos(x) times frac(cos(x))(sin(x))#
Finally, let's apply another standard trigonometric identity; #cot(x) = frac(cos(x))(sin(x))#:
#= cos(x) times cot(x)#
#= cos(x) cot(x)# ## Q.E.D.
