If $\tan \theta +\sin \theta =m$ and $\tan \theta -\sin \theta =n$, then which one of the following relations is correct?

Let us calculate $m^2-n^2$ using the algebraic identity $a^2-b^2=(a-b)(a+b)$:
$$m+n=(\tan \theta +\sin \theta )+(\tan \theta -\sin \theta )=2\tan \theta$$

$$m-n=(\tan \theta +\sin \theta )-(\tan \theta -\sin \theta )=2\sin \theta$$

Multiplying these two expressions together:

$$m^2-n^2=(m+n)(m-n)=(2\tan \theta )(2\sin \theta )=4\tan \theta \sin \theta \text{— (Equation 1)}$$

Now let us compute the product $mn$:

$$mn=(\tan \theta +\sin \theta )(\tan \theta -\sin \theta )={\tan }^2\theta -{\sin }^2\theta$$

$${\tan }^2\theta -{\sin }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }-{\sin }^2\theta ={\sin }^2\theta \left(\frac{1}{{\cos }^2\theta }-1\right)$$

$$mn={\sin }^2\theta ({\sec }^2\theta -1)={\sin }^2\theta {\tan }^2\theta$$

Taking the square root on both sides:

$$\sqrt{mn}=\sin \theta \tan \theta$$

Multiplying by $4$:

$$4\sqrt{mn}=4\tan \theta \sin \theta \text{— (Equation 2)}$$

Comparing Equation 1 and Equation 2:

$$m^2-n^2=4\sqrt{mn}$$