How do you verify the identity #cotalpha+tanalpha=cscalphasecalpha#?

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Answer 1 · 2017-01-19T06:29:09

See proof below

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

#cotalpha=cosalpha/sinalpha#

#tanalpha=sinalpha/cosalpha#

#cscalpha=1/sinalpha#

#secalpha=1/cosalpha#

Therefore,

#LHS=cotalpha+tanalpha#

#=cosalpha/sinalpha+sinalpha/cosalpha#

#=(cos^2alpha+sin^2alpha)/(sinalphacosalpha)#

#=1/(sinalphacosalpha)#

#=cscalphasecalpha#

#=RHS#

#QED#