Question

1+ tan x / 1+ cot x = sec x / csc x

Answer 1

LHS \[\frac{1+tanx}{1+cotx}=\huge {\dfrac{1+\frac{sinx}{cosx}}{1+\frac{cosx}{sinx}}}\]
\[\huge \dfrac{\frac{cosx+sinx}{cosx}}{\frac{sinx+cosx}{sinx}}=\frac{sinx}{cosx}=tanx\]

Answer 2

RHS \[\huge \frac{secx}{cscx}= \dfrac{\frac{1}{cosx}}{\frac{1}{sinx}}=\frac{sinx}{cosx}=tanx=LHS\]