How do you prove #csc B/cos B - cosB/sin B = tan B#?

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Answer 1 · 2015-11-26T03:04:35

It is explained below

#=(csc B sin B- cos^2 B)/(sin B cos B)#

=(1-cos^2 B)/(sin B cos B)#

= #(sin^2 B)/(sin B cos B)#
= #(sin B)/cos B= tan B#

Answer 2 · 2015-11-26T03:15:35

#cscB/cosB-cosB/sinB=tanB#

LS:
#cscB/cosB-cosB/sinB#

#=(1/sinB)/cosB-cosB/sinB#

#=(1/sinB-:cosB)-cosB/sinB#

#=(1/sinB*1/cosB)-cosB/sinB#

#=1/(sinBcosB)-cosB/sinB#

#=(1-cosB(cosB))/(sinBcosB)#

#=(1-cos^2B)/(sinBcosB)#

#=sin^2B/(sinBcosB)#

#=(sinBsinB)/(sinBcosB)#

#=(color(red)cancelcolor(black)(sinB)sinB)/(color(red)cancelcolor(black)(sinB)cosB)#

#=tanB#

#:.#, LS = RS.