Question #fbd74
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"Which following pairs of atoms, have a lower electron affinity? a) Ca,K b) I,F c) Li, Ra. I seriously don't know anything about electron affinity all ik that it can buy another element"
#(cos^2x + tan^2x - 1)/tan^2x =sin^2x#
Work on left side:
#(cos^2x + tan^2x - (cos^2x+sin^2x))/tan^2x#
#cos^2x# cancels, leaving:
#(tan^2x-sin^2x)/tan^2x#
Change tan to #sin/cos#:
#((sin^2x/cos^2x)-(Sin^2x/1))/(sin^2x/cos^2x)#
Multiply #(Sin^2x/1)# by #(cos^2x/cos^2x)# to get a common denominator:
#((sin^2x/cos^2x)-((Sin^2xcos^2x)/cos^2x))/(sin^2x/cos^2x)#
#((Sin^2x-sin^2xcos^2x)/cos^2x)/(sin^2x/cos^2x)#
#cos^2x# cancels:
#(sin^2x-sin^2xcos^2x)/sin^2x#
Factor the numerator:
#(sin^2x(1-cos^2x))/sin^2x#
#sin^2x# cancels:
#1-cos^2x# which is also #sin^2x# = right side
#LHS=(cos^2x + tan^2x - 1)/tan^2x#
#=( tan^2x - (1-cos^2x))/tan^2x#
#=( tan^2x - sin^2x)/tan^2x#
#= tan^2x/tan^2x - sin^2x/tan^2x#
#=1 - sin^2x/(sin^2x/cos^2x)#
#=1 - cos^2x=sin^2x=RHS#
