Question Number 102178 by dw last updated on 07/Jul/20

Answered by 1549442205 last updated on 07/Jul/20

Putting x=tanα,y=tanβ we have  (√(1+x^2 ))=(√(1+tan^2 α))=(1/(cosα)),(√(1+y^2 ))=(1/(cosβ))  x(√(1+y^2 ))+y(√(1+x^2 ))=((tamα)/(cosβ))+((tanβ)/(cosα))=((sinα+sinβ)/(cosαcosβ)).Hence,  LHS=(1/(cosαcosβ))(√(1+(((sinα+sinβ)/(cosαcosβ)))^2 ))+(1/(cosαcosβ))(√(1+(((sinα−sinβ)/(cosαcosβ)))^2 ))  =(1/(cosαcosβ))(√(cos^2 αcos^2 β+(sinα+sinβ)^2 ))+(1/(cosαcosβ))(√(cos^2 αcos^2 β+(sinα−sinβ)^2 ))  we have :cos^2 αcos^2 β=(1−sin^2 α)(1−sin^2 β)  =1+sin^2 αsin^2 β−sin^2 α−sin^2 β,so  =(√(cos^2 αcos^2 β+(sinα+sinβ)^2  ))=(√(1+sin^2 αsin^2 β+2sinαsinβ))  =(√((sinαsinβ+1)^2 )) =1+sinαsinβ  Similarly,=(√(cos^2 αcos^2 β+(sinα−sinβ)^2 ))=1−sinαsinβ  Therefore,LHS=(1/(cosαcosβ))[(1+sinαsinβ)−(1−sinαsinβ)]  =((2sinαsinβ)/(cosαcosβ))=2tanαtanβ=2xy  Thus,LHS=RHS(q.e.d)

Commented bydw last updated on 07/Jul/20

Thank you! Your solution is very good.

Commented bydw last updated on 16/Jul/20

but I did not understand !    (1/(cosαcosβ))[(1+sinαsinβ)−(1−sinαsinβ)]=    (1/(cosαcosβ))[(1+sinαsinβ)+(1−sinαsinβ)