Question 32892
tan^3x-1/tanx-1 = sec^2x+tan x..USING....A^3-B^3=(A-B)(A^2+AB+B^2)
LHS=[{TAN(X)-1}{TAN^2(X)+TAN(X)+1}]/[(TAN(X)-1]=SEC^2(X)+TAN(X)=RHS...SINCE 1+TAN^2(X)=SEC^2(X)