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

Hint: This question belongs to the topic of trigonometry. We should know the formulas of trigonometry to solve this type of question easily. In solving this question, first we will multiply the term which is outside the parenthesis or bracket with the term which is inside the parenthesis or bracket. After that, we will do the further equation.

Complete step by step answer:
Let us solve this question.
In this question, we have asked to simplify the expression \[\cos t\left( \sec t-\cos t \right)\] or we can say that we have to solve the term \[\cos t\left( \sec t-\cos t \right)\] and make it simple.
The term which we have to solve is \[\cos t\left( \sec t-\cos t \right)\] can also be written as by multiplying the term inside the parenthesis with the term outside the parenthesis or bracket. So, we can write
\[\cos t\left( \sec t-\cos t \right)=\cos t\times \sec t-\cos t\times \cos t\]
As we know that the cos function is the inverse of sec function or we can say \[\cos t\] can also be written as \[\dfrac{1}{\sec t}\]. So, we can write the above equation as
\[\Rightarrow \cos t\left( \sec t-\cos t \right)=\dfrac{1}{\sec t}\times \sec t-\cos t\times \cos t\]
The equation can also be written as
\[\Rightarrow \cos t\left( \sec t-\cos t \right)=1-{{\cos }^{2}}t\]
As we know that \[{{\sin }^{2}}t+{{\cos }^{2}}t=1\]. The equation \[{{\sin }^{2}}t+{{\cos }^{2}}t=1\] can also be written as \[{{\sin }^{2}}t=1-{{\cos }^{2}}t\].
Hence, using the formula \[{{\sin }^{2}}t=1-{{\cos }^{2}}t\] we can write the above equation as
\[\Rightarrow \cos t\left( \sec t-\cos t \right)={{\sin }^{2}}t\]
So, now we have found the simplified value of \[\cos t\left( \sec t-\cos t \right)\]. The simplified value of \[\cos t\left( \sec t-\cos t \right)\] is \[{{\sin }^{2}}t\].

Note: As one can see that this question is from the topic of trigonometry. So, we should have a better knowledge in that topic.. The formulas that can be used in this type of question are in the following:
\[{{\sin }^{2}}t+{{\cos }^{2}}t=1\]
\[\sec t=\dfrac{1}{\cos t}\]
\[\csc t=\dfrac{1}{\sin t}\]
\[\tan t=\dfrac{\sin t}{\cos t}\]
The above identities and formulas should be kept remembered to solve this type of question easily.