How do you multiply \[{\left( {\cos x + \sec x} \right)^2}\]and write the product in simplest form?

Hint: Here in this question, we have to find the product of 2 binomials. The binomial is one form of algebraic expression. The given expression in the form of square first we write the two times the same expression and then we use the arithmetic operation that is multiplication and then we simplify.

Complete step-by-step solution:
The binomial concept will come under the topic of algebraic expressions. The algebraic expression is a combination of variables and constant. The alphabets are known as variables and the numerals are known as constants. In algebraic expression or equation, we have 3 types namely, monomial, binomial and polynomial.
A polynomial equation with two terms joined by the arithmetic operation + or – is called as binomial equation.
Now let us consider the given expression \[{\left( {\cos x + \sec x} \right)^2}\]
It can be written as
\[ \Rightarrow \,\left( {\cos x + \sec x} \right)\left( {\cos x + \sec x} \right)\]
 two binomial and they are \[\left( {\cos x + \sec x} \right)\], and \[\left( {\cos x + \sec x} \right)\] which two binomials are same
Now we have to multiply the binomials, to multiply the binomials we use multiplication. The multiplication is one of the arithmetic operations.
Now we multiply the above 2 binomials we get
\[ \Rightarrow \,\left( {\cos x + \sec x} \right)\left( {\cos x + \sec x} \right)\]
Here dot represents the multiplication. First, we multiply the first two terms of the above equation
\[ \Rightarrow \cos x\left( {\cos x + \sec x} \right) + \sec x\left( {\cos x + \sec x} \right)\]
On multiplying we get
\[ \Rightarrow {\cos ^2}x + \cos x\sec x + \cos x\sec x + {\sec ^2}x\]
On simplification we have
\[ \Rightarrow {\cos ^2}x + 2\cos x\sec x + {\sec ^2}x\]
Hence, we have multiplied the two same binomial and obtained the solution for the question
Therefore, we have \[{\left( {\cos x + \sec x} \right)^2} = {\cos ^2}x + 2\cos x\sec x + {\sec ^2}x\]

Note: We can also solve the given question by using the standard algebraic formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]. Here \[a = \cos x\]and \[\;b = \sec x\]. On substituting the values in the formula, we get
\[ \Rightarrow \,{\left( {\cos x + \sec x} \right)^2} = \,{\left( {\cos x} \right)^2} - 2\left( {\cos x} \right)\left( {\sec x} \right) + {\left( {\sec x} \right)^2}\]
\[ \Rightarrow \,{\left( {\cos x + \sec x} \right)^2} = \,{\cos ^2}x + 2\cos x\sec x + {\sec ^2}x\]