Which expression is equal to $\sqrt{\dfrac{1+\sin A}{1-\sin A}}$ for an acute angle $A$?

Underlying idea: rationalise the expression inside the square root using $1-\sin^2 A=\cos^2 A$. $$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sqrt{\frac{(1+\sin A)^2}{1-\sin^2 A}}=\sqrt{\frac{(1+\sin A)^2}{\cos^2 A}}.$$ Since $A$ is acute, $\cos A>0$, so this becomes $$\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}=\sec A+\tan A.$$ Option A is tempting because $(\sec A-\tan A)(\sec A+\tan A)=1$, so it is actually the reciprocal, not the given expression. Option B uses cosec and cot, which arise from dividing by $\sin A$, not $\cos A$.