Prove that
tanA. +. cotA
------- ------- =1+secAcosec
1-cotA. 1-tanA
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Answer:
tanA/(1-cotA) +cotA/(1-tanA)
=tanA/(1–1/tanA) +cotA/(1-tanA)
=tan^2A/(tanA-1) +cotA/(1-tanA)
=-tan^2A/(1-tanA) +cotA/(1-tanA)
=(-tan^2A+cotA)/(1-tanA)
=(-tan^2A+1/tanA)/(1-tanA)
=(-tan^3A+1)/tanA(1-tanA)
=(1-tan^3A)/tanA(1-tanA)
=(1-tanA)(1+tanA+tan^2A)/tanA(1-tanA)
=(1+tanA+tan^2A)/tanA
=(1+tan^2A+tanA)/tanA
=(sec^2A+tanA)/tanA
=sec^2A/tanA +tanA/tanA
=1/cos^2AtanA+1
=cosA/cos^2AsinA+1
=1/cosAsinA+1
=secAcosecA+1
Step-by-step explanation:
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