Question #7f9eb

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Answer 1 · 2017-12-07T07:19:59

#Sin(x)/(1-cos(x)) + (1-cos(x))/sin(x)#

#=(Sin(x)(1+cos(x)) )/((1-cos(x))(1+cos(x)) ) + (1-cos(x))/sin(x)#

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

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

#=(1+cos(x)) /sin(x)+ (1-cos(x))/sin(x)#

#=(1+cos(x)+1-cos(x))/sin(x)#

#=2/sin(x)=2csc(x)#

Answer 2 · 2017-12-07T07:24:07

The answer is #=2cscx#

We need

#sin^2x+cos^2x=1#

#1/sinx=cscx#

#(a-b)^2=a^2+b^2-2ab#

Therefore,

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

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

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

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

#=2/sinx#

#=2cscx#