Question

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

Answer 1

Hint: We first simplify the numerator and the denominator part of $1-{{\sec }^{2}}\theta $ separately. We use the identities ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$. We also know that the terms $\sec \theta $ and $\cos \theta $ are inverse of each other which gives $\sec \theta \times \cos \theta =1$. For both parts we get the value of the simplification as equal to $-{{\tan }^{2}}\theta $.

Complete step-by-step solution:
We have been given a trigonometric expression of $1-{{\sec }^{2}}\theta $.
For the expression we have the identity of ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$.
We subtract ${{\sec }^{2}}\theta $ from the both sides of the equation ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$.
$\begin{align}
  & {{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1 \\
 & \Rightarrow {{\sec }^{2}}\theta -{{\tan }^{2}}\theta -{{\sec }^{2}}\theta =1-{{\sec }^{2}}\theta \\
 & \Rightarrow -{{\tan }^{2}}\theta =1-{{\sec }^{2}}\theta \\
\end{align}$
We get $1-{{\sec }^{2}}\theta =-{{\tan }^{2}}\theta $.
Therefore, the simplified form of $1-{{\sec }^{2}}\theta $ is $-{{\tan }^{2}}\theta $.
We can also solve it using the inverse theorem.
We know that the terms $\sec \theta $ and $\cos \theta $ are inverse of each other.
So, $\sec \theta \times \cos \theta =1$ as $\sec \theta =\dfrac{1}{\cos \theta }$.
We multiply ${{\cos }^{2}}\theta $ to both the terms of $1-{{\sec }^{2}}\theta $. We also need to keep ${{\cos }^{2}}\theta $ divided in the denominator.
From the exponent theorem we get that for two numbers $a$ and $b$ we have ${{a}^{2}}\times {{b}^{2}}={{\left( ab \right)}^{2}}$.
Then we multiply ${{\cos }^{2}}\theta $ with ${{\sec }^{2}}\theta $ and get ${{\cos }^{2}}\theta \times {{\sec }^{2}}\theta $. The value of ${{\cos }^{2}}\theta \times {{\sec }^{2}}\theta $ can be replaced as 1 as $\cos \theta \sec \theta =1$.
So, ${{\cos }^{2}}\theta \left( 1-{{\sec }^{2}}\theta \right)={{\cos }^{2}}\theta -{{\cos }^{2}}\theta \times {{\sec }^{2}}\theta ={{\cos }^{2}}\theta -1$.
Now we use the theorem \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\]. This gives \[{{\cos }^{2}}\theta -1=-{{\sin }^{2}}\theta \].
So, $1-{{\sec }^{2}}\theta =\dfrac{{{\cos }^{2}}\theta \left( 1-{{\sec }^{2}}\theta \right)}{{{\cos }^{2}}\theta }=\dfrac{-{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }=-{{\tan }^{2}}\theta $.

Note: We need to remember that the final terms are square terms. The identities \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] and $\cos \theta \sec \theta =1$ are valid for any value of $\theta $. The division of the fraction part only gives $-{{\tan }^{2}}\theta $ as the solution.