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