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A+B=90^(@)impliesB=90^(@)-A :.1+(tanA)/(tanB)=1+(tanA)/(tan(90^(@)-A)) =1+(tanA)/(cotA)=1+tanA.tanA =1+tan^(2)A=sec^(2)A. Hence 1+(tanA)/(tanB)=sec^(2)A.
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Bengali] If A+B=90^(@) then prove that 1+(tanA)/(tanB)=sec^(2)A.