How do you verify #sinx/(1-cosx) + (1-cosx)/sinx = 2cscx#?

Question

55078 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T20:51:10

See Below

#LHS: sin x/(1-cos x) +(1-cosx)/sin x#

#=(sinx*sinx+(1-cosx)(1-cosx))/(sinx(1-cos x))#->common denominator

#=(sin^2 x+1-2cosx+cos^2x)/(sinx(1-cosx)#

#=(sin^2 x+cos^2x+1-2cosx)/(sinx(1-cosx))#

#=(1+1-2cosx)/(sinx(1-cosx))#

#=(2-2cosx)/(sinx(1-cosx))#

#=(2(1-cosx))/(sinx(1-cosx))#

#=(2(cancel(1-cosx)))/(sinx cancel((1-cosx)))#

#=2/sinx#

#=2*1/sinx#

#=2 csc x#

#=RHS#