If #sec \theta + tan \theta = m#, how do you show that #(m^2 -1)/(m^2 + 1) = sin \theta#?

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Answer 1 · 2017-12-03T07:58:57

See the proof below

We need

#tan^2theta+1=sec^2theta#

Therefore,

#LHS=((sectheta+tantheta)^2-1)((sectheta+tantheta)^2+1)#

#=(sec^2theta+tan^2theta+2secthetatantheta-1)/(sec^2theta+tan^2theta+2secthetatantheta+1)#

#=(2tan^2theta+2secthetatantheta)/(2sec^2theta+2secthetatantheta)#

#=(2tanthetacancel(tantheta+sectheta))/(2secthetacancel(tantheta+sectheta))#

#=tantheta/sectheta#

#=sintheta/cancelcostheta*cancelcostheta#

#=sintheta#

#=RHS#

#QED#