How can you rewrite this trigonometric expression as an algebraic expression: $\cos \left( {{{\sin }^{ - 1}}\left( u \right) - {{\cos }^{ - 1}}\left( v \right)} \right)$ ?

Hint: In order to solve this particular question, we take ${\sin ^{ - 1}}\left( u \right)$ as $\alpha $ and ${\cos ^{ - 1}}\left( v \right)$ as $\beta $ . Then, our expression becomes $\cos \left( {\alpha - \beta } \right)$ , we expand this expression according to the formula $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$ and replace the appropriate sin and cos values with $u$ and $v$ to get our required answer.

Complete step-by-step solution:
In the given question, we are asked to express the trigonometric function $\cos \left( {{{\sin }^{ - 1}}\left( u \right) - {{\cos }^{ - 1}}\left( v \right)} \right)$ into algebraic function. To express them in algebraic form means to simplify them in variables. In order to do so, let us do the following-
Let us take ${\sin ^{ - 1}}\left( u \right) = \alpha $ and ${\cos ^{ - 1}}\left( v \right) = \beta $
Now if we take our inverse functions to the opposite sides, we get:
$u = \sin \alpha $ and $v = \cos \beta $ → let this be equation (1)
Thus, now our expression becomes: $\cos \left( {\alpha - \beta } \right)$
And as we know that $\cos \left( {A - B} \right) = \cos A\cos B + \sin A\sin B$
Therefore, $\cos \left( {\alpha - \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta $ → let us call this as equation (2)
Now, we know that ${\sin ^2}\alpha + {\cos ^2}\alpha = 1$
Therefore $\cos \alpha $ can also be written as $\sqrt {1 - {{\sin }^2}\alpha } $
Similarly, we can also write this expression as ${\sin ^2}\beta + {\cos ^2}\beta = 1$
Thus, $\sin \beta $ can also be written as $\sqrt {1 - {{\cos }^2}\beta } $
Now placing these alternate values of $\cos \alpha $ and $\sin \beta $ in equation (2), we get:
$\cos \left( {\alpha + \beta } \right) = \sqrt {1 - {{\sin }^2}\alpha } \cos \beta - \sin \alpha \sqrt {1 - {{\cos }^2}\beta } $
If we recall equation (1), we will find that $u = \sin \alpha $ and $v = \cos \beta $
Thus, $\cos \left( {\alpha + \beta } \right) = \sqrt {1 - {u^2}} v - u\sqrt {1 - {v^2}} $
Rearranging the above expression properly, we get:
$\cos \left( {\alpha + \beta } \right) = v \times \sqrt {1 - {u^2}} - u \times \sqrt {1 - {v^2}} $

Thus $\cos \left( {{{\sin }^{ - 1}}\left( u \right) - {{\cos }^{ - 1}}\left( v \right)} \right)$ can be expressed algebraically as $v \times \sqrt {1 - {u^2}} - u \times \sqrt {1 - {v^2}} $

Note: We have solved this sum using a substitution method to substitute the appropriate trigonometric values of sin and cos. Trigonometry is a branch of mathematics which deals with triangles. There are many trigonometric formulas that establish a relation between the lengths and angles of respective triangles. In trigonometry, we use a right-angled triangle to find ratios of its different sides and angles such as sine, cosine, tan, and their respective inverse like cosec, sec, and cot. Some common formulas of trigonometric identities are:
$\sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}$ , where perpendicular is the side containing the right angle in a right-angled triangle and hypotenuse is the side opposite to the perpendicular.
$\cos \theta = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}$ , where base is the side containing the perpendicular and hypotenuse
$\tan \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}$