How to Prove Below Inverse Sin How to prove below equation .
$$     \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}.$$
I am not able to prove the above equation how can we prove it.
As we know 
$  \sin^{-1} = y$ if $\sin y = x$ where $-1\leq x\leq 1, -\pi/2 \leq y \leq \pi/2.$
 A: Remark: The question has changed from $\sin^{-1}(5/13)+\sin^{-1}(3/5)$ to 
$\sin^{-1}(3/5)+\sin^{-1}(8/17)$. There is no good reason to change the answer below, since the process is basically the same. 
Hint: Use the "sum" formula $\sin(a+b)=\sin a \cos b+\cos a\sin b$.
If $a$ is the angle between $0$ and $\pi/2$ whose sine is $\frac{5}{13}$, then the cosine of $a$ is $\frac{12}{13}$, since always $\cos^2 x+\sin^2 x=1$.
If $b$ is the angle between $0$ and $\pi/2$ whose sine is $\frac{3}{5}$, then the cosine of $b$ is $\frac{4}{5}$.
You will also need to show that the sum on the left is $\lt \pi/2$.
Remark: The notation $\sin^{-1} t$ is a fairly frequent source of confusion. It means the angle between $-\pi/2$ and $\pi/2$ whose sine is $t$. It does not mean $\frac{1}{\sin t}$. So for example $\sin^{-1}(1)=\pi/2$ and $\sin^{-1}(1/2)=\pi/6$. 
A: Let  $\displaystyle\sin^{-1}\frac35= A$
$\displaystyle\implies (i)\sin A=\frac35$ and $(ii)-\frac\pi2< A<\frac\pi2 $ as the Principal value of sine inverse lies in that range $\displaystyle\implies\cos A>0 \implies \cos A=+\sqrt{1-\left(\frac35\right)^2}=+\frac45$
Similarly for $\displaystyle\sin^{-1}\frac5{13}= B$
Use $\sin(A+B)=\sin A\cos B+\cos A\sin B$
From this, $$\sin^{-1}x+\sin^{-1}y=$$
$$ \begin{cases} \sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) &\mbox{if } x^2+y^2\le1\text{ or if } x^2+y^2>1\text{ and } xy<0  \\ \pi-\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) & \mbox{if } x^2+y^2>1\text{ and }x,y>0 \\  -\pi-\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) & \mbox{if } x^2+y^2>1\text{ and }x,y<0 \end{cases} $$
