If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$ 
If $\tan\theta +\sin\theta=m$ and $m^2 -n^2=4\sqrt{mn}$ so prove that $\tan\theta-\sin\theta=n$

I found the similar question in Quora. There was slightly a mistake.
$$(\tan\theta+\sin\theta)^2-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\tan\theta\sin\theta-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\sqrt{\tan^2\theta-\sin^2\theta}-n^2=4\sqrt{mn}$$
$$(\tan\theta-\sin\theta)^2+4\sqrt{m(\tan\theta-\sin\theta)}-n^2-4\sqrt{mn}=0$$
I just said that "we have to take $4\sqrt{m(\tan\theta-\sin\theta)}=4\sqrt{mn}$ to satisfy $\tan\theta-\sin\theta=n$ but I don't like it, I just want direct derivation."

And in quora they just took $4\sqrt{m}$ common and wrote $$(\tan\theta-\sin\theta)^2+4\sqrt{m}(\sqrt{\tan\theta-\sin\theta-n})-n^2=0$$
which is totally wrong

cause $\sqrt{\tan\theta-\sin\theta-n}\neq\sqrt{\tan\theta-\sin\theta}-\sqrt{n}$
 A: Let's call $x=\tan\theta-\sin\theta$ then
$$(\tan\theta-\sin\theta)^2+4\sqrt{m(\tan\theta-\sin\theta)}-n^2-4\sqrt{mn}=0$$
becomes
\begin{align}
x^2+4\sqrt{m}\sqrt{x}-n^2-4\sqrt{m}\sqrt{n}&=0\\
(x^2-n^2)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(x-n)(x+n)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(\sqrt{x}-\sqrt{n})(\sqrt{x}+\sqrt{n})(x+n)+4\sqrt{m}(\sqrt{x}-\sqrt{n})&=0\\
(\sqrt{x}-\sqrt{n})\Big((\sqrt{x}+\sqrt{n})(x+n)+4\sqrt{m}\Big)&=0
\end{align}
and since the second term is strictly positive $\sqrt{x}=\sqrt{n}\iff x=n$ as we wanted.
A: Assume that $\theta\in (0,\pi/2)$ and hence $\tan\theta,\,\sin\theta>0$.
If $w=\sin\theta$, then $\tan\theta=\frac{w}{\sqrt{1-w^2}}$ and
$$
m=w+\frac{w}{\sqrt{1-w^2}}=\frac{w+w\sqrt{1-w^2}}{\sqrt{1-w^2}}
$$
Now
$$
F(n)=m^2-n^2-4\sqrt{mn}, \quad n\ge 0, 
$$
is strictly decreasing and continuous $F(0)=m^2>0$ and $F(\infty)=-\infty$. Hence, it vanishes for a unique $n$. It suffices to show that
$$
F(\tan\theta-\sin\theta)=F\left(\frac{w-w\sqrt{1-w^2}}{\sqrt{1-w^2}}\right)=0
$$
which is straight-forward.
