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

2 Answers
Nov 26, 2015

It is explained below

Explanation:

=(csc B sin B- cos^2 B)/(sin B cos B)=cscBsinBcos2BsinBcosB

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

= (sin^2 B)/(sin B cos B)sin2BsinBcosB
= (sin B)/cos B= tan BsinBcosB=tanB

Nov 26, 2015

cscB/cosB-cosB/sinB=tanBcscBcosBcosBsinB=tanB

LS:
cscB/cosB-cosB/sinBcscBcosBcosBsinB

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

=(1/sinB-:cosB)-cosB/sinB=(1sinB÷cosB)cosBsinB

=(1/sinB*1/cosB)-cosB/sinB=(1sinB1cosB)cosBsinB

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

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

=(1-cos^2B)/(sinBcosB)=1cos2BsinBcosB

=sin^2B/(sinBcosB)=sin2BsinBcosB

=(sinBsinB)/(sinBcosB)=sinBsinBsinBcosB

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

=tanB

:., LS = RS.