How can #(1+cosx)/(sinx) + (sinx)/(1+cosx)=2cscx#?

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Answer 1 · 2017-05-04T07:07:25

We need

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

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

#cscx=1/sinx#

Therefore,

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

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

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

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

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

#=2/sinx#

#=2cscx#

#=RHS#

#QED#