How do you simplify the expression \[1-{{\sec }^{2}}\theta ?\]

Hint: We are given an expression as \[1-{{\sec }^{2}}\theta \] and we are asked to simplify it. To simplify it means we have to arrange it in the simplest form or we try to represent it by simple function or so. To do so we will need to know how \[{{\sec }^{2}}\theta \] and other relations are connected to each other. With these connections only we can solve and simplify and so we will learn the various ratios and relations.

Complete step by step answer:
We are given a problem in which we have to simplify \[1-{{\sec }^{2}}\theta .\] To do so we will learn how the rationals are connected. We will need the connection between \[\sin \theta ,\cos \theta ,\sec \theta ,\operatorname{cosec}\theta \] and then \[\sin \theta ,\cos \theta ,\tan \theta .\] So we will start by learning them and then we will use them. So, as we know,
\[\sec \theta =\dfrac{1}{\cos \theta }.....\left( i \right)\]
We also have that
\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
So, using this we can see that \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta .\]
And lastly, we know that \[\tan \theta \] is formed by \[\sin \theta \] and \[\cos \theta \] and it is given as \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }.\]
Now using all this we will simplify our problem. We have \[1-{{\sec }^{2}}\theta \] as \[\sec \theta =\dfrac{1}{\cos \theta }\] so we get,
\[{{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }\]
Using this above, we get,
\[1-{{\sec }^{2}}\theta =1-\dfrac{1}{{{\cos }^{2}}\theta }\]
On simplifying, we get,
\[\Rightarrow 1-{{\sec }^{2}}\theta =\dfrac{{{\cos }^{2}}\theta -1}{{{\cos }^{2}}\theta }\]
As \[1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta \] so \[{{\cos }^{2}}\theta -1=-{{\sin }^{2}}\theta \]
Using this above, we get,
\[\Rightarrow 1-{{\sec }^{2}}\theta =\dfrac{-{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\]
Taking – sign out and put square above the whole term, we get,
\[\Rightarrow 1-{{\sec }^{2}}\theta =-{{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}\]
\[\Rightarrow 1-{{\sec }^{2}}\theta =-{{\tan }^{2}}\theta \left[ \text{As }\dfrac{\sin \theta }{\cos \theta }=\tan \theta \right]\]
So, we get,
\[\Rightarrow 1-{{\sec }^{2}}\theta =-{{\tan }^{2}}\theta \]

Note: We can also simplify such problems directly by using the identity given as \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta .\] We will subtract \[{{\tan }^{2}}\theta \] both sides and we get,
\[1+{{\tan }^{2}}\theta -{{\tan }^{2}}\theta ={{\sec }^{2}}\theta -{{\tan }^{2}}\theta \]
So,
\[\Rightarrow 1={{\sec }^{2}}\theta -{{\tan }^{2}}\theta \]
We will now subtract \[{{\sec }^{2}}\theta \] and we will get
\[\Rightarrow 1-{{\sec }^{2}}\theta ={{\sec }^{2}}\theta -{{\sec }^{2}}\theta -{{\tan }^{2}}\theta \]
So, simplifying, we get,
\[\Rightarrow 1-{{\sec }^{2}}\theta =-{{\tan }^{2}}\theta \]
Hence, the result is correct.