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

We have,
$\mathrm { LHS } = \frac { \tan A } { 1 - \cot A } + \frac { \cot A } { 1 - \tan A }$
$\Rightarrow \quad \mathrm { LHS } = \frac { \tan A } { 1 - \frac { 1 } { \tan A } } + \frac { \frac { 1 } { \tan A } } { 1 - \tan A }$
$\Rightarrow \quad \text { LHS } = \frac { \tan A } { \frac { \tan A -1 } { \tan A } } + \frac { 1 } { \tan A ( 1 - \tan A ) }$
$\Rightarrow \quad \mathrm { LHS } = \frac { \tan ^ { 2 } A } { \tan A - 1 } + \frac { 1 } { \tan A ( 1 - \tan A ) }$
$\Rightarrow \quad \text { LHS } = \frac { \tan ^ { 2 } A } { \tan A - 1 } - \frac { 1 } { \tan A ( \tan A - 1 ) }$
$\Rightarrow \quad \mathrm { LHS } = \frac { \tan ^ { 3 } A - 1 } { \tan A ( \tan A - 1 ) }$ [Taking LCM]
$\Rightarrow \quad \mathrm { LHS } = \frac { ( \tan A - 1 ) \left( \tan ^ { 2 } A + \tan A + 1 \right) } { \tan A ( \tan A - 1 ) }$ [$\because$ a³ - b³ = ( a - b )(a² + ab + b²)]
$\Rightarrow \quad \mathrm { LHS } = \frac { \tan ^ { 2 } A + \tan A + 1 } { \tan A }$
$\Rightarrow \quad \mathrm { LHS } = \frac { \tan ^ { 2 } A } { \tan A } + \frac { \tan A } { \tan A } + \frac { 1 } { \tan A }$
$\Rightarrow$ LHS = tanA + 1 + cotA         [ since (1/tanA) =cotA ].
= (1 + tanA + cotA)
$\therefore \quad \frac { \tan A } { 1 - \cot A } + \frac { \cot A } { 1 - \tan A }$ = 1 + tanA + cotA ...........(1)

Now, 1 + tanA + cotA
  = 1 + $\frac { \sin A } { \cos A } + \frac { \cos A } { \sin A }$
 = 1 + $\frac { \sin ^ { 2 } A + \cos ^ { 2 } A } { \sin A \cos A }$  = 1 + $\frac { 1 } { \sin A \cos A }$ [$\because$Sin²A + Cos² A = 1 ] 
= 1 + cosecAsecA
So, 1 + tanA + cotA = 1+ cosecAsecA.......(2)

From (1) and (2), we obtain
$\frac { \tan A } { 1 - \cot A } + \frac { \cot A } { 1 - \tan A }$ = 1 + tanA + cotA = 1 + cosecAsecA