Question

Prove that: tanA/I-cottA+cotA/1-tanA=secAcocesA+1-​

Answer 1

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