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

Hint: In this question we have to convert the given trigonometric expression into an algebraic expression, and can do this by first considering the inverse trigonometry functions as variables and using the trigonometry identities $\cos A = \sqrt {1 - {{\sin }^2}A} $ and $\sin A = \sqrt {1 - {{\cos }^2}A} $ we can find the cos and sin values for the respective sin and cos values, we will get the expression in cos and substituting the values in the identity $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$, we will get the required result.

Complete step-by-step answer:
Given trigonometric expression is $\cos \left( {{{\sin }^{ - 1}}\left( u \right) + {{\cos }^{ - 1}}\left( v \right)} \right)$,
Let us consider${\sin ^{ - 1}}u = A$ and one of the angles of a right triangle.
Then, set the hypotenuse of the triangle equal to 1. So, the adjacent side of A is u, using the Pythagorean formula, the opposite side is $\sqrt {1 - {u^2}} $ . Base on this triangle,
$\sin A = \dfrac{{opp}}{{hyp}} = \dfrac{u}{1}$ and$\cos A = \dfrac{{adj}}{{hyp}} = \dfrac{{\sqrt {1 - {u^2}} }}{1}$,
$\sin A = u$and$\cos A = \sqrt {1 - {u^2}} $,
Let us consider${\cos ^{ - 1}}v = B$ and one of the angles of a right triangle.
Then, set the hypotenuse of the triangle equal to 1. So, the adjacent side of B is v. Using the Pythagorean formula, the opposite side is $\sqrt {1 - {v^2}} $. Base on this triangle,
$\cos B = \dfrac{{adj}}{{hyp}} = \dfrac{v}{1}$ and$\sin A = \dfrac{{opp}}{{hyp}} = \dfrac{{\sqrt {1 - {v^2}} }}{1}$,
$\cos B = v$ and $\sin B = \sqrt {1 - {v^2}} $,
Now given expression is$\cos \left( {{{\sin }^{ - 1}}\left( u \right) + {{\cos }^{ - 1}}\left( v \right)} \right)$,
Now substituting the values we get,$\cos \left( {A + B} \right)$,
By using the formula $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$,
Now substituting the values in the formula we get, here $\sin A = u$,$\cos A = \sqrt {1 - {u^2}} $,$\cos B = v$ and $\sin B = \sqrt {1 - {v^2}} $, we get,
$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}u + {{\cos }^{ - 1}}v} \right) = v \cdot \sqrt {1 - {u^2}} - u \cdot \sqrt {1 - {v^2}} $,

$\therefore $The algebraic expression when the trigonometric expression $\cos \left( {{{\sin }^{ - 1}}u + {{\cos }^{ - 1}}v} \right)$ is converted will be equal to $v \cdot \sqrt {1 - {u^2}} - u \cdot \sqrt {1 - {v^2}} $.

Note:
An algebraic expression is an expression involving numbers, parentheses, operation signs and numerals that becomes a number when numbers are substituted for the numerals. For example, $2x + 5$ is an expression but $ + , \times $ is not.
Trigonometric expression is an expression which includes the trigonometric functions of a variable.
For example:$1 - {\sin ^2}x$.