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

2 Answers
Nov 26, 2015

It is explained below

Explanation:

=(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

Nov 26, 2015

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.