How do you prove #sinx/(1-cosx) + (1-cosx)/sinx = 2csc x#?

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Answer 1 · 2018-08-10T17:02:32

Please see below.

We take ,

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

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

#LHS#=#(1-cos^2x)/(sinx(1-cosx))+(1-cosx)/sinxto[because sin^2x+cos^2x#=#1]#

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

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

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

#LHS=2/sinx#

#LHS=2cscx#

#LHS=RHS#