1.

If `A+B=90^(@)` then prove that `1+(tanA)/(tanB)=sec^(2)A.`

Answer» `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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