Prove that : ${{\sin }^{-1}}\dfrac{3}{5}+{{\sin }^{-1}}\dfrac{8}{17}={{\sin }^{-1}}\dfrac{77}{85}$.

Hint: We will be using the concept of inverse trigonometric functions. We will be using the formula of ${{\sin }^{-1}}x+{{\sin }^{-1}}y$.

Complete step by step answer:
Now, we have to prove that ${{\sin }^{-1}}\dfrac{3}{5}+{{\sin }^{-1}}\dfrac{8}{17}={{\sin }^{-1}}\dfrac{77}{85}$.
We will be taking the left hand side of the equation and prove it to be equal to the right hand side.
Now, taking L.H.S we have,
${{\sin }^{-1}}\dfrac{3}{5}+{{\sin }^{-1}}\dfrac{8}{17}...........\left( 1 \right)$
We know that ${{\sin }^{-1}}x+{{\sin }^{-1}}y$ is $={{\sin }^{-1}}\left( x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right).........\left( 2 \right)$
Where $x,y\ge 0\ and\ {{x}^{2}}+{{y}^{2}}\le 1$.
We will be using (2) to solve (1), but first we have to check whether $x,y\ge 0\ and\ {{x}^{2}}+{{y}^{2}}\le 1$.
Comparing (1) and (2) we have,
$\begin{align}
  & x=\dfrac{3}{5}>0 \\
 & y=\dfrac{8}{17}>0 \\
 & Also, \\
 & {{x}^{2}}+{{y}^{2}}=\dfrac{9}{25}+\dfrac{64}{289} \\
 & =\dfrac{9\times 289+64\times 25}{25\times 289} \\
 & =\dfrac{4201}{25\times 289} \\
 & \approx 0.5 \\
\end{align}$
So, this shows that equation (1) satisfy both the condition as $x,y\ge 0\ and\ {{x}^{2}}+{{y}^{2}}=0.5$ is less than 1.
Now, using equation (2),
$\begin{align}
  & {{\sin }^{-1}}\left( \dfrac{3}{5} \right)+{{\sin }^{-1}}\left( \dfrac{8}{17} \right)={{\sin }^{-1}}\left( \dfrac{3}{5}\sqrt{1-\dfrac{{{8}^{2}}}{{{17}^{2}}}}+\dfrac{8}{17}\sqrt{1-\dfrac{{{3}^{2}}}{{{5}^{2}}}} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{3}{5}\sqrt{\dfrac{225}{172}}+\dfrac{8}{17}\sqrt{\dfrac{16}{{{5}^{2}}}} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{3}{5}\times \sqrt{\dfrac{{{15}^{2}}}{{{17}^{2}}}}+\dfrac{8}{17}\sqrt{\dfrac{{{4}^{2}}}{{{5}^{2}}}} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{3\times 15}{5\times 17}+\dfrac{8}{17}\times \dfrac{4}{5} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{9}{17}+\dfrac{32}{85} \right) \\
 & ={{\sin }^{-1}}\left( \dfrac{77}{85} \right) \\
\end{align}$
L.H.S = R.H.S
Hence Proved.

Note: These types of questions are calculation and formula based. So, remembering the formulas of trigonometric functions and checking calculations is a must.