How do you simplify the expression \[\left( 1-\cos x \right)\left( 1+\sec x \right)\left( \cos x \right)\]?

Hint: Assume the given expression as ‘E’. Now, leave the term \[\left( 1-\cos x \right)\] as it is and consider the product of \[\left( 1+\sec x \right)\] and \[\cos x\] in the first step. Use the conversion: - \[\sec x=\dfrac{1}{\cos x}\] to simplify the product. Now, convert the expression into the form \[\left( a+b \right)\left( a-b \right)\] and use the algebraic identity: - \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\] for further simplification. In the final step of solution use the trigonometric identity: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] to get the answer.

Complete step by step solution:
Here, we have been provided with the trigonometric expression \[\left( 1-\cos x \right)\left( 1+\sec x \right)\left( \cos x \right)\] and we are asked to simplify it. Here, we have to use some basic algebraic and trigonometric identities to get the simplified. Let us assume this expression as E, so we have,
\[\Rightarrow E=\left( 1-\cos x \right)\left( 1+\sec x \right)\left( \cos x \right)\]
Now, we know that secant function is the reciprocal of cosine function which is given mathematically as \[\sec \theta =\dfrac{1}{\cos \theta }\], so now leaving the term \[\left( 1-\cos x \right)\] and considering the product of other two terms we get,
\[\begin{align}
  & \Rightarrow E=\left( 1-\cos x \right)\left( \cos x+\sec x\cos x \right) \\
 & \Rightarrow E=\left( 1-\cos x \right)\left( \cos x+\dfrac{1}{\cos x}\times \cos x \right) \\
 & \Rightarrow E=\left( 1-\cos x \right)\left( \cos x+1 \right) \\
\end{align}\]
We can write the above expression as: -
\[\Rightarrow E=\left( 1-\cos x \right)\left( \cos x+1 \right)\]
Clearly, the above expression is of the form \[\left( a+b \right)\left( a-b \right)\] if we will consider a = 1 and \[b=\cos x\], so using the algebraic identity: - \[\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)\], we get,
\[\begin{align}
  & \Rightarrow E={{1}^{2}}-{{\left( \cos x \right)}^{2}} \\
 & \Rightarrow E=1-{{\cos }^{2}}x \\
\end{align}\]
Now, in any right-angle triangle we have \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] any value of the angel \[\theta \], so using this identity, we get,
\[\Rightarrow E={{\sin }^{2}}x\]

Hence, the simplified value of the given expression is \[{{\sin }^{2}}x\].

Note: One may note that it is not necessary to consider the product of \[\left( 1+\sec x \right)\] and \[\cos x\] first you may consider the product of any two terms first. The simplified value will be the same. The most important thing you have to remember is certain algebraic identities and three basic trigonometric identities given as: - \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], \[1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta \] and \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \]. Without using these algebraic identities it will be very difficult to solve the question.