How do you find \[\cot \left( {{{\sin }^{ - 1}}\left( {\dfrac{p}{q}} \right)} \right)?\]

Hint: In the given problem first we need to concentrate on different trigonometric ratios given in the problem, here it is cotangent and sine. We are trying to convert sine inverse into cotangent inverse so that they will get canceled out and the required value will get obtained.

Formula used:
Pythagoras theorem, \[{h^2} = {p^2} + {b^2}\] where, (\[h = \]Hypotenuse, \[p = \]perpendicular, \[b = \]base)
Cosec\[\theta \]\[ = \left( {\dfrac{h}{p}} \right)\] where, (\[h = \]Hypotenuse, \[p = \]perpendicular)

Complete step by step solution:
First try to convert sine inverse into cot inverse, for that let sine inverse be ‘\[\theta \]’ and write it as follows
\[ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{p}{q}} \right)\]
Taking sine inverse to left hand side, we get
\[ \Rightarrow \sin \theta = \left( {\dfrac{p}{q}} \right)\]
Comparing above equation with \[\sin \theta = \dfrac{p}{h}\] where, (\[h = \]Hypotenuse, \[p = \]perpendicular), we get
\[ \Rightarrow p = p\] and \[h = q\]
Therefore, we got following equation,
\[ \Rightarrow \]\[\cos ec\theta = \left( {\dfrac{q}{p}} \right)\]
So, we can replace \[\sin \theta \] with \[\cos ec\theta \] and write it as following,
\[ \Rightarrow \]\[\cos ec\theta = \left( {\dfrac{q}{p}} \right)\]
Taking \[\cos ec\] to the right-hand side so that it will get converted into inverse form,
\[ \Rightarrow \]\[\theta = \cos e{c^{ - 1}}\left( {\dfrac{q}{p}} \right)\]
Now, the expression given in problem can be replaced by above equation, we get
\[ \Rightarrow \]\[\cot \left( {\cos e{c^{ - 1}}\left( {\dfrac{q}{p}} \right)} \right)\]
Now, we are trying to convert the above expression into cotangent inverse form.
For doing so, we have to calculate \[\cot \theta \] in the form of \[q\] and \[p\]. \[\cot \theta \] can be written as,
\[ \Rightarrow \cot \theta = \dfrac{b}{p}\] where, (\[b = \]base, \[p = \]perpendicular)
By using Pythagoras theorem, it can also be written as following,
\[ \Rightarrow \cot \theta = \dfrac{{\sqrt {{h^2} - {p^2}} }}{p}\]
As we know by above conclusion that \[p = p\] and \[h = q\], it can also be written as,
\[ \Rightarrow \cot \theta = \dfrac{{\sqrt {{q^2} - {p^2}} }}{p}\]
Taking \[p\] common from numerator, we get
\[ \Rightarrow \cot \theta = \dfrac{{\sqrt {{p^2}\left( {\dfrac{{{q^2}}}{{{p^2}}} - 1} \right)} }}{p}\]
Taking \[p\] outside from under-root numerator, we get
\[ \Rightarrow \cot \theta = \dfrac{{p\sqrt {\left( {\dfrac{{{q^2}}}{{{p^2}}} - 1} \right)} }}{p}\]
\[p\] will get cancel out with denominator,
\[ \Rightarrow \cot \theta = \sqrt {\left( {\dfrac{{{q^2}}}{{{p^2}}} - 1} \right)} \]
Simplifying above equation by taking L.C.M, we get
\[ \Rightarrow \cot \theta = \sqrt {\left( {\dfrac{{{q^2} - {p^2}}}{{{p^2}}}} \right)} \]
Taking \[p\] outside from under-root denominator, we get
\[ \Rightarrow \cot \theta = \dfrac{{\sqrt {{q^2} - {p^2}} }}{p}\]
Taking \[\cot \]right hand side so that it will get converted into inverse form as follows,
\[ \Rightarrow \theta = {\cot ^{ - 1}}\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}\]
It can also be written as following,
\[ \Rightarrow \theta = {\sin ^{ - 1}}\left( {\dfrac{p}{q}} \right) = \]\[{\cot ^{ - 1}}\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}\]
So, we are going to replace the expression given in problem with the equation obtained above,
\[ \Rightarrow \]\[\cot \left( {{{\cot }^{ - 1}}\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}} \right)\]
\[\cot \]will get cancel out and we obtain the final required value as following,
\[ \Rightarrow \]\[\left( {\dfrac{{\sqrt {{q^2} - {p^2}} }}{p}} \right)\]

Note:
We have to make relations between different trigonometric ratios so that one trigonometric ratio can be easily converted into another trigonometric ratio with the help of Pythagoras theorem.