Evaluate ${\cot ^{ - 1}}\left( {\dfrac{3}{4}} \right) + {\sin ^{ - 1}}\left( {\dfrac{5}{{13}}} \right) = $ ?
A.${\sin ^{ - 1}}\left( {\dfrac{{63}}{{65}}} \right)$
B.${\sin ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)$
C.${\sin ^{ - 1}}\left( {\dfrac{{65}}{{68}}} \right)$
D.${\sin ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)$

Hint: We have to find the addition of two inverse trigonometric functions. The functions are not added like the normal trigonometric functions. We will first convert the first term of expression which is in the form of cot inverse into the sin inverse form and then add the sin inverse function together to get our final answer.

Formula Used:
The formula to add the sine inverse functions are given below:
${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}[x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} ]$
The relation of $\sin \theta $ with $\cot \theta $:
\[\sin \theta {\text{ }} = {\text{ }}\dfrac{{1{\text{ }}}}{{\sqrt {1{\text{ }} + {\text{ co}}{{\text{t}}^2}\;\theta } }}\]

Complete step-by-step answer:
We are given an expression containing the inverse trigonometric functions as:
${\cot ^{ - 1}}\left( {\dfrac{3}{4}} \right) + {\sin ^{ - 1}}\left( {\dfrac{5}{{13}}} \right)$
As we know there is no direct formula to add these two inverse trigonometric function, so we will first convert the ${\cot ^{ - 1}}$ to ${\sin ^{ - 1}}$, for this let:
\[\;co{t^{ - 1}}\dfrac{3}{4} = {\text{ }}\theta \]
\[cot{\text{ }}\theta {\text{ }} = {\text{ }}\dfrac{3}{4}{\text{ }}\]
Finding the $\sin \theta $ from this equation,
We know:
\[\sin \theta {\text{ }} = {\text{ }}\dfrac{{1{\text{ }}}}{{\sqrt {1{\text{ }} + {\text{ co}}{{\text{t}}^2}\;\theta } }}\]
Thus we can write:
\[sin{\text{ }}\theta {\text{ }} = \dfrac{1}{{{\text{ }}\sqrt {1{\text{ }} + {\text{ }}\dfrac{9}{{16}}} }}{\text{ }}\]
$ \Rightarrow \sin \theta = \dfrac{4}{5}$
\[ \Rightarrow \theta = {\sin ^{ - 1}}\dfrac{4}{5}\]
Since we took \[\;co{t^{ - 1}}\dfrac{3}{4} = {\text{ }}\theta \] the ${\cot ^{ - 1}}$term can be replaced by \[{\sin ^{ - 1}}\dfrac{4}{5}\]
The original expression will now be written as:
\[ \Rightarrow {\cot ^{ - 1}}\;\left( {\dfrac{3}{4}} \right){\text{ }} + {\text{ si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{5}{{13}}} \right) = {\text{ si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{4}{5}} \right){\text{ }} + {\text{ si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{5}{{13}}} \right)\]
Now we will add the two sine inverse using the formula:
${\sin ^{ - 1}}x + {\sin ^{ - 1}}y = {\sin ^{ - 1}}[x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} ]$
Thus we will add as:
\[ \Rightarrow {\text{si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{4}{5}} \right) + {\text{si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{5}{{13}}} \right) = s{\text{i}}{{\text{n}}^{ - 1}}\;\left[ {\left( {\dfrac{4}{5}} \right) \times \sqrt {1-\left( {\dfrac{{{5^2}}}{{{{13}^2}}}} \right)} {\text{ }} + \left( {\dfrac{5}{{13}}} \right) \times \sqrt {1-\left( {\dfrac{{{4^2}}}{{{5^2}}}{\text{ }}} \right)} } \right]\]
\[ \Rightarrow {\text{si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{4}{5}} \right) + {\text{si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{5}{{13}}} \right) = {\text{si}}{{\text{n}}^{ - 1}}\;\left[ {\left( {\dfrac{4}{5}} \right) \times \sqrt {1-\left( {\dfrac{{25}}{{169}}} \right)} + \left( {\dfrac{5}{{13}}} \right) \times \sqrt {1-\left( {\dfrac{{16}}{{25}}{\text{ }}} \right)} } \right]\]
Solving this further we get right hand side as:
\[ \Rightarrow {\text{ si}}{{\text{n}}^{ - 1}}\;\left[ {\left( {\dfrac{{4{\text{ }}}}{5}} \right){\text{ }} \times {\text{ }}\left( {\dfrac{{12}}{{13}}} \right){\text{ }} + {\text{ }}\left( {\dfrac{5}{{13}}} \right){\text{ }} \times {\text{ }}\left( {\dfrac{3}{5}{\text{ }}} \right)} \right]\]
\[ \Rightarrow {\text{si}}{{\text{n}}^{ - 1}}\;\left( {\dfrac{{48{\text{ }} + {\text{ }}15}}{{65}}} \right)\]
Upon solving we can thus write as:
\[ = {\text{ }}si{n^{ - 1}}\;\left( {\dfrac{{63}}{{65}}} \right)\]
which corresponds to option 1 as given in the question.
Thus ${\cot ^{ - 1}}\left( {\dfrac{3}{4}} \right) + {\sin ^{ - 1}}\left( {\dfrac{5}{{13}}} \right) = {\text{ }}si{n^{ - 1}}\;\left( {\dfrac{{63}}{{65}}} \right)$

Note: Just like the above question, it is to be taken into the account by the student that there are no direct formulas to add two different inverse trigonometric functions to each other, instead, the student must learn to convert quickly one inverse trigonometric function to the other so that the addition is facilitated. The conversion can be done by first taking the given inverse trigonometric angle equal to $\theta $ and then solving to achieve different trigonometric functions for the same $\theta $, and then inverting it.