Question #1d07e

Question

191 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-25T16:36:30

See proof below

We need

#cotalpha=cosalpha/sinalpha#

#cscalpha=1/sinalpha#

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

Therefore,

#LHS=1+(cot^2alpha)/(1+cscalpha)#

#=1+(cos^2alpha/sin^2alpha)/(1+1/sinalpha)#

#=1+(cos^2alpha/sin^2alpha)/((sinalpha+1)/sinalpha)#

#=1+cos^2alpha/(sinalpha(sinalpha+1))#

#=(sin^2alpha+sinalpha+cos^2alpha)/(sinalpha(sinalpha+1))#

#=cancel(1+sinalpha)/(sinalphacancel(1+sinalpha))#

#=1/sinalpha#

#=cscalpha#

#=RHS#

#QED#

I hope that this is helpful!