LHS
[tex]\\ \sf\longmapsto \dfrac{tanA+secA-1}{tanA-secA+1}[/tex]
As we know
[tex]\\ \boxed{\sf sec^2A-tan^2A=1}[/tex]
[tex]\\ \sf\longmapsto \dfrac{tanA+secA-(sec^2A-tan^2A)}{tanA-secA+1}[/tex]
[tex]\\ \sf\longmapsto \dfrac{tanA+secA\cancel{(1-secA+tanA)}}{\cancel{(1-secA+tanA)}}[/tex]
[tex]\\ \sf\longmapsto tanA+secA[/tex]
[tex]\\ \sf\longmapsto \dfrac{sinA}{cosA}+\dfrac{1}{cosA}[/tex]
[tex]\\ \sf\longmapsto \dfrac{1+sinA}{cosA}[/tex]
[tex]\\ \sf\longmapsto RHS[/tex]
Hence proved
