Question #897a1

Question

986 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-17T18:15:21

see below

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

Left Side :#=(1-cosx)/sinx + sinx/(1-cosx) #

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

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

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

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

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

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

#=2/sinx#

#:.=# Right Side