Answer: \(\dfrac1{1-\sin\theta}-\dfrac1{1+\sin\theta}=2\tan\theta\operatorname{sec}\theta\).

Start with the left-hand side:

\(\dfrac1{1-\sin\theta}-\dfrac1{1+\sin\theta}\).

Use the common denominator \((1-\sin\theta)(1+\sin\theta)\):

\(\dfrac{(1+\sin\theta)-(1-\sin\theta)}{(1-\sin\theta)(1+\sin\theta)}\).

The numerator simplifies to \(2\sin\theta\), so

\(\dfrac{2\sin\theta}{1-\sin^2\theta}\).

Using \(1-\sin^2\theta=\cos^2\theta\), this becomes

\(\dfrac{2\sin\theta}{\cos^2\theta}\).

Now

\(\dfrac{\sin\theta}{\cos^2\theta}=\dfrac{\sin\theta}{\cos\theta}\cdot\dfrac1{\cos\theta}=\tan\theta\operatorname{sec}\theta.\)

Therefore

\(\dfrac1{1-\sin\theta}-\dfrac1{1+\sin\theta}=2\tan\theta\operatorname{sec}\theta.\)