How do you simplify ${\tan ^4}\theta + 2{\tan ^2}\theta + 1$?

Hint: In this problem we have given a trigonometric equation where the highest power of the given equation is $4$. Moreover the given equation is in the form of a perfect square. And we are asked to simplify the given trigonometric equation. This problem can be simplified by using some trigonometric identities. So by using some trigonometric identities and Pythagorean identity we are going to solve this problem.

Formula used: ${\tan ^2}\theta + 1 = {\sec ^2}\theta $
${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step-by-step solution:
Given is, ${\tan ^4}\theta + 2{\tan ^2}\theta + 1$
If we see, ${\tan ^4}\theta + 2{\tan ^2}\theta + 1$ is a perfect square, we can remember the formula $\left( {{x^2} + {y^2}} \right) = {x^2} + {y^2} + 2xy$,
Now we use this formula in the given equation then we get,
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\left( {{{\tan }^2}\theta + 1} \right)^2} - - - - - (1)$
Now, the formula ${\sin ^2}\theta + {\cos ^2}\theta = 1$ is the Pythagorean identity.
By using Pythagorean identity, we write \[\dfrac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\cos }^2}\theta }} = \dfrac{1}{{{{\cos }^2}\theta }}\]
Also we know that the one of the trigonometric identity, ${\tan ^2}\theta + 1 = {\sec ^2}\theta $
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\tan ^4}\theta + {\tan ^2}\theta + {\tan ^2}\theta + 1$,
Now, considering the last two terms of right hand side, we get
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\tan ^4}\theta + {\tan ^2}\theta + ({\tan ^2}\theta + 1)$
\[ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\tan ^4}\theta + {\tan ^2}\theta + {\sec ^2}\theta \],
Next we take ${\tan ^2}\theta $ as common in the right hand side, we get
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\tan ^2}\theta \left( {{{\tan }^2}\theta + 1} \right) + {\sec ^2}\theta $
Again substitute, ${\tan ^2}\theta + 1 = {\sec ^2}\theta $, we get
$ \Rightarrow {\tan ^2}\theta + 1 = {\sec ^2}\theta = {\tan ^2}\theta \left( {{{\sec }^2}\theta } \right) + {\sec ^2}\theta $,
Now take ${\sec ^2}\theta $ as common, we get
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\sec ^2}\theta \left( {{{\tan }^2}\theta + 1} \right)$, using${\tan ^2}\theta + 1 = {\sec ^2}\theta $, we get
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\sec ^2}\theta \times {\sec ^2}\theta $
Let us multiply the term and we get,
$ \Rightarrow {\tan ^4}\theta + 2{\tan ^2}\theta + 1 = {\sec ^4}\theta $
Therefore simplifying ${\tan ^4}\theta + 2{\tan ^2}\theta + 1$ we get ${\sec ^4}\theta $.

Hence the required answer is ${\sec ^4}\theta $.

Note: Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined.
If we want to reduce the steps to solve this problem then that is possible. Already we know that the given equation is in the form of a perfect square.
Also in equation (1) we expressed it. Then by using the identity ${\tan ^2}\theta + 1 = {\sec ^2}\theta $ we get ${({\sec ^2}\theta )^2}$ this implies ${\sec ^4}\theta $. By this way we can reduce the steps of this problem.