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If angle C of triangle ABC is 90^0, then...

If angle C of triangle ABC is `90^0,` then prove that `tanA+tanB=(c^2)/(a b)` (where, `a , b , c ,` are sides opposite to angles `A , B , C ,` respectively).

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Draw `DeltaABC " with " angleC=90^@`. We have
`tanA+tanB=a/b+b/a=(a^2+b^2)/(ab)=c^2/(ab)`
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