We known that$$\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right).$$
Now we derive the above formula. Let$$
\tan^{-1}a=\theta _1 \implies \tan\theta_1=a,\\
\tan^{-1}b=\theta _2 \implies \tan\theta_2=b,\\
\theta _1+\theta _2 = \tan^{-1}a+ \tan^{-1}b,\\
\tan(\theta _1+\theta _2) = \tan(\tan^{-1}a+\tan^{-1}b)·
\tan(\theta _1+\theta _2)\\
=\frac{ \tan(\tan^{-1}a)+\tan( \tan^{-1}b)}{1- \tan(\tan^{-1}a)· \tan(\tan^{-1}b)},\\
(\theta _1+\theta _2) =\tan^{-1}\left(\frac {a+b}{1-ab}\right).
$$
Therefore, $\tan^{-1} a +\ tan^{-1}b=\tan^{-1}\left(\dfrac{a+b}{1-ab}\right)$.
For all $\theta _1$ and $\theta _2$ such that $0 \le \theta _1,\theta _2 \lt 90$, in the similar way I try to find $\sin^{-1}a +\sin^{-1}b$. Let$$
\sin^{-1}a=\theta_1, \quad \sin^{-1}b=\theta_2,\\
\sin(\theta_1+\theta_2)=\sin(\sin^{-1}a +\sin^{-1}b),\\
\sin(\theta_1+\theta_2)=\sin(\sin^{-1}a)·\cos(sin^{-1}b) +\sin(\sin^{-1}b)·\cos(sin^{-1}a),\\
\theta_1+\theta_2=\sin(\sin^{-1}a)·\cos(sin^{-1}b) +\sin(\sin^{-1}b)·\cos(sin^{-1}a),\\
\sin^{-1}a+\sin^{-1}b=\sin(\sin^{-1}a).cos(sin^{-1}b) +\sin(\sin^{-1}b).cos(sin^{-1}a),\\
\sin^{-1}a+\sin^{-1}b=\sin(a)·\cos(\sin^{-1}b) +\sin(b)·\cos(\sin^{-1}a).$$
It cannot be write in the form only using $a$ and $b$.
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A·\tan B}.$$ It contains only $\tan$ terms. But $\sin(A+B)=\sin A·\cos B+\cos A·\sin B$, which contains both $\sin$ and $\cos$ terms.
Is there any formula for $\sin^{-1}a+\sin^{-1}b$?
Best Answer
Assume that $\arcsin(a)+\arcsin(b)\in\left[-\frac\pi2,\frac\pi2\right]$. Then, since$$\sin\left(\arcsin(a)+\arcsin(b)\right)=a\sqrt{1-b^2}+b\sqrt{1-a^2},$$we have$$\arcsin(a)+\arcsin(b)=\arcsin\left(a\sqrt{1-b^2}+b\sqrt{1-a^2}\right).$$