Prove $\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$ (part of sum) I'm looking for a specific clarification for a part of the solution in proving the following identity.
$\tan^{-1}m+\tan^{-1}n=\cos^{-1}\frac{1-mn}{\sqrt{1+m^2}\sqrt{1+n^2}}$
Here I'm taking, $\theta=\tan^{-1}m;$ $-\pi/2<\theta<\pi/2$
so I get $\tan\theta=m$---(1) 
I need to find $\sin \theta  $and  $\cos \theta$ in terms of m
By Trigonometric identity I can easily derive, $\cos \theta$
$\tan^2\theta+1=\sec^2\theta$
$\cos^2\theta=\frac{1}{m^2+1}$
$\cos\theta=+\sqrt\frac{1}{m^2+1}$ ( here only plus due to range of $\theta$)
Now if I deduce $\sin\theta$ from,
$\sin^2\theta+\cos^2\theta=1$

I get, $\sin\theta=\pm\sqrt\frac{m^2}{m^2+1}$ ( I have to take $\pm$ because of range of $\theta$)

But if I deduce $\sin\theta$ from (1)

I get, $\sin\theta=\frac{m}{\sqrt {m^2+1}}$

Which of the following method is correct to find $\sin\theta$? Please help me. Thank you!
P.S. I'm not interested in the solution. What I need to know is how to find $\sin\theta$
 A: Using this
We need $\tan^{-1}m+\tan^{-1}n\ge0$  to admit the equality
$$\tan^{-1}m+\tan^{-1}n\ge0\iff\tan^{-1}m>-\tan^{-1}n=\tan^{-1}(-n)$$
$$\iff m\ge-n\iff m+n\ge0$$
Now use Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$
Then if $\tan^{-1}x=y,-\dfrac\pi2<y<\dfrac\pi2,\cos y>0$
$x=\tan y\implies\cos y=\dfrac1{\sqrt{1+x^2}}$
$\tan^{-1}x=\text{sign of}(x)\cdot\cos^{-1}\dfrac1{\sqrt{1+x^2}}$
See also : Proving that $\cos^{-1}\frac{4}{5}+\cos^{-1}\frac{12}{13}=\cos^{-1}\frac{33}{65}$
A: I don't have means to put up the image for my plot so bear with me.
Take a right-angled triangle $ABC$ such that AB is the height $=m$ and BC is the base $=1$. $\widehat{ABC}=90^\circ$ and $\widehat{ACB}=\tan^{-1} m$
Let $AC=y$. Draw a line $AE$ perpendicular to $AC$ at $A$ equal to $ny$. $\widehat{EAC}=90^\circ$ and $\widehat{ACE}=\tan^{-1} \dfrac{ny}{y}=\tan^{-1} n$. Additionally, let $EC=z$
Extend $AB$ past $A$. Draw a perpendicular from E to that line and name that intersection $D$. $\widehat{EAD}=\widehat{ACB}$. This is because of the right angle $\widehat{EAC}$. Hence $\triangle ABC \sim \triangle EDA$. This equality follows:
$\dfrac{AB}{AC}=\dfrac{ED}{EA}$
$\dfrac{m}{y}=\dfrac{ED}{ny}$
$ED=mn$
Draw a perpendicular to $BC$ at $C$ and extend $DE$ past $E$. Label the point where these lines meet as $F$. We have created a rectangle in $DBCF$, thus $DF=BC=1$:
$EF+DE=1$
$EF=1-mn$
Additionally, $DF \parallel BC$ hence $\widehat{ECB}=\widehat{CEF} \text{  Alternate angles}$
$\widehat{ECB}=\widehat{ACB}+\widehat{ECA}=\tan^{-1} m+\tan^{-1} n$
$\widehat{CEF}=\cos^{-1} \dfrac{EF}{EC}$
$\dfrac{EF}{EC}=\dfrac{1-mn}{z}$
Remember now all the right-angled triangles we drew.
$z^2=(ny)^2+y^2=y^2(n^2+1)$
$y^2=m^2+1$
$z^2=(n^2+1)(m^2+1)$
$z=\sqrt{(n^2+1)(m^2+1)}$
$\dfrac{EF}{EC}=\dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}$
$\widehat{CEF}=\cos^{-1} \dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}$
But $\widehat{ECB}=\widehat{CEF}$, hence:
$\boxed{\tan^{-1} m+\tan^{-1} n=\cos^{-1} \dfrac{1-mn}{\sqrt{(n^2+1)(m^2+1)}}}$
Edit: the formula kinda breaks down when $m$ and $n$ don't have the same sign i.e $n \lt 0 \lt m$ for example. But they won't be that different. For example, using $-\sqrt{3}$ and $\frac{1}{\sqrt{3}}$ yields $-30^\circ$ and $30^\circ$ on the left and right respectively. But they do have the same cosine value meaning they're not too distinct.
