Question

tanA/(1-cotA)+cotA/(1-tanA)=1+secAcosecA

Answer 1

cotA=1/tanA
tanA=1/cotA
tanA/1-cotA+cotA/1-tanA=1+secAcosecA
tanA/1-1/tanA + cotA/1-1/cotA
tanA/tanA-1/tanA + cotA/cotA-1/cotA
tanA²/tanA-1+1/tanA/1-1/tanA
tanA²/tanA-1 - 1/tanA(tanA-1)
taking lcm on both sides
tanA³-1/tanA(tanA-1)
using identity a³+b³ =(a-b)(a²+ab+b²)
(tanA-1)(tan²A+tanA+1)/tanA(tanA-1)
(tan²A +tanA+ 1)/tanA
tan²A/tanA+tanA/tanA+1/tanA
tanA+1+cot A
tanA=sinA/cos A and cosA/sinA=cotA
sinA/cosA+1+cosA/sinA
(sin²A+cos²A+cosAsinA)/cosAsinA
sin²A+cos²A=1
1/cosAsinA + 1
1/cosA=secA and 1/sinA=cosecA 
secAcosecA+1 
hence lhs = rhs so proves