If ${\sin ^{ - 1}}\left( {\dfrac{3}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\sin ^{ - 1}}\left( C \right)$, then $C = $
1. $\dfrac{{65}}{{56}}$
2. $\dfrac{{24}}{{65}}$
3. $\dfrac{{16}}{{65}}$
4. $\dfrac{{56}}{{65}}$

Hint: Here, in the given question, we are given an equation of inverse trigonometric functions and we need to find the value of $C$. At first, we will convert the inverse function of $\cos $ into inverse function of $\sin $ using ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ identity.

Formula used:
We will apply ${{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right\}$ identity to get the value of $C$.


Complete step by step solution:
Given that, ${{\sin }^{-1}}\left( \frac{3}{5} \right)+{{\cos }^{-1}}\left( \frac{12}{13} \right)={{\sin }^{-1}}\left( C \right)$
Let us first convert ${{\cos }^{-1}}$ function into ${{\sin }^{-1}}$
The inverse functions of the fundamental trigonometric functions are known as inverse trigonometric functions. $\sin ^{-1} x=\theta$ is a possible conversion for the fundamental trigonometric function$x$. In this case, $x$ can be expressed as a whole number, a decimal, a fraction, or an exponent. We have $(1 / 2)$ for $\theta=30^{\circ}$, where is a number between $0^{\circ}$and $90^{\circ}$. Inverse trigonometric function formulas can be created from any trigonometric formula.
Let $A={{\cos }^{-1}}\left( \frac{12}{13} \right)$
$\Rightarrow {{\sin }^{-1}}\left( \frac{3}{5} \right)+A={{\sin }^{-1}}\left( C \right)\,\,\,\,\,.......\left( i \right)$
We have, $A={{\cos }^{-1}}\left( \frac{12}{13} \right)$
$\Rightarrow \cos A=\frac{12}{13}$
As we know that ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Therefore, we have
$\Rightarrow {{\sin }^{2}}A+{{\left( \frac{12}{13} \right)}^{2}}=1$
Transfer ${{\left( \frac{12}{13} \right)}^{2}}$ to the R.H.S.
$\Rightarrow {{\sin }^{2}}A=1-{{\left( \frac{12}{13} \right)}^{2}}$
$\Rightarrow {{\sin }^{2}}A=1-\frac{144}{169}$
On taking L.C.M., we get
$\Rightarrow {{\sin }^{2}}A=\frac{169-144}{169}$
$\Rightarrow {{\sin }^{2}}A=\frac{25}{169}$
Take square root on both the sides
$\Rightarrow \sin A=\sqrt{\frac{25}{169}}=\frac{5}{13}$
$\Rightarrow A={{\sin }^{-1}}\left( \frac{5}{13} \right)$
Now, substitute the value of $A={{\sin }^{-1}}\left( \frac{5}{13} \right)$ in equation $\left( i \right)$
$\Rightarrow {{\sin }^{-1}}\left( \frac{3}{5} \right)+{{\sin }^{-1}}\left( \frac{5}{13} \right)={{\sin }^{-1}}\left( C \right)$
AS We know that ${{\sin }^{-1}}x+{{\sin }^{-1}}y={{\sin }^{-1}}\left\{ x\sqrt{1-{{y}^{2}}}+y\sqrt{1-{{x}^{2}}} \right\}$. Therefore, we get
\[\Rightarrow {{\sin }^{-1}}\left\{ \frac{3}{5}\sqrt{1-{{\left( \frac{5}{13} \right)}^{2}}}+\frac{5}{13}\sqrt{1-{{\left( \frac{3}{5} \right)}^{2}}} \right\}={{\sin }^{-1}}\left( C \right)\]
\[\Rightarrow {{\sin }^{-1}}\left\{ \frac{3}{5}\sqrt{1-\frac{25}{169}}+\frac{5}{13}\sqrt{1-\frac{9}{25}} \right\}={{\sin }^{-1}}\left( C \right)\]
On taking L.C.M., we get
\[\Rightarrow {{\sin }^{-1}}\left\{ \frac{3}{5}\sqrt{\frac{169-25}{169}}+\frac{5}{13}\sqrt{\frac{25-9}{25}} \right\}={{\sin }^{-1}}\left( C \right)\]
\[\Rightarrow {{\sin }^{-1}}\left\{ \frac{3}{5}\sqrt{\frac{144}{169}}+\frac{5}{13}\sqrt{\frac{16}{25}} \right\}={{\sin }^{-1}}\left( C \right)\]
We can write the above written equation as
\[\Rightarrow {{\sin }^{-1}}\left\{ \frac{3}{5}\sqrt{{{\left( \frac{12}{13} \right)}^{2}}}+\frac{5}{13}\sqrt{{{\left( \frac{4}{5} \right)}^{2}}} \right\}={{\sin }^{-1}}\left( C \right)\]
\[\Rightarrow {{\sin }^{-1}}\left( \frac{3}{5}\times \frac{12}{13}+\frac{5}{13}\times \frac{4}{5} \right)={{\sin }^{-1}}\left( C \right)\]
On multiplication of terms, we get
\[\Rightarrow {{\sin }^{-1}}\left( \frac{36}{65}+\frac{20}{65} \right)={{\sin }^{-1}}\left( C \right)\]
On taking L.C.M., we get
\[\Rightarrow {{\sin }^{-1}}\left( \frac{56}{65} \right)={{\sin }^{-1}}\left( C \right)\]
$\therefore C=\frac{56}{65}$
Hence, the value of $C$ is $\frac{56}{65}$.
Therefore, the correct option is 4.

Note: The trigonometric functions sine, cosine, tangent, cosecant, secant, and cotangent really execute the opposite operation, which is what the inverse trigonometric functions do. We are aware that the right angle triangle is particularly amenable to trig functions.
If the responses are the smallest values possible, then ${{\sin }^{-1}}x,{{\cos }^{-1}}x,{{\tan }^{-1}}x$
 etc. signify angles or real numbers whose sine is x, cosine is x, and tangent is x.