Question

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

Answer 1

In this question the trigonometric expression is given, for simplify this expression we will use the basic trigonometric identity. And there are some trigonometric identities as given below.
$\Rightarrow \dfrac{{\sin x}}{{\cos x}} = \tan x$
$\Rightarrow \sin x = \dfrac{1}{{\csc x}}$
$\Rightarrow \cos x = \dfrac{1}{{\sec x}}$
$\Rightarrow \tan x = \dfrac{1}{{\cot x}}$

Complete step by step answer:
In this question, the word trigonometric identity is used. First, we know about trigonometric identities.
The trigonometric identities are defined as the equations that are true for right-angle triangles.
We determine the trigonometric identities by using the right-angle triangle. Its angle is $x$. Here $b$ is the length of the base, $p$ is the height and $h$ is the length of the hypotenuse.
There are three main functions in trigonometry: sin, cos, and tan.
Then we write these functions according to the right angle triangle.

$$\sin x = \dfrac{p}{h} \\\ \cos x = \dfrac{b}{h} \\\ \tan x = \dfrac{p}{b} \\\$$

Then,
$\tan x$ is written as below.
$\Rightarrow \dfrac{{\sin x}}{{\cos x}} = \dfrac{{p/h}}{{b/h}} = \dfrac{p}{b} = \tan x$
Then,
$\Rightarrow \tan x = \dfrac{{\sin x}}{{\cos x}}$
And another trigonometric function is written as below (by using right angle triangle).

$$\csc x = \dfrac{h}{p} = \dfrac{1}{{\sin x}} \\\ \sec x = \dfrac{h}{b} = \dfrac{1}{{\cos x}} \\\ \cot x = \dfrac{b}{p} = \dfrac{1}{{\tan x}} \\\$$

We can also write it as,

$$\sin x = \dfrac{p}{h} = \dfrac{1}{{\cos ecx}} \\\ \cos x = \dfrac{b}{h} = \dfrac{1}{{\sec x}} \\\ \tan x = \dfrac{p}{b} = \dfrac{1}{{\cot x}} \\\$$

Now, in the question, the trigonometric form is given which is simplified as below.
$1 - {\sec ^2}x$
We know that according to the right angle triangle $\sec x = \dfrac{h}{b}$.
Put this value in the above equation.
Then, the above equation is written as.
$\Rightarrow 1 - \dfrac{{{h^2}}}{{{b^2}}}$
Now, we simplify the above equation.
Then,
$\Rightarrow \dfrac{{{b^2} - {h^2}}}{{{b^2}}}$
We know that, according to Pythagoras theorem.

$$\Rightarrow {h^2} = {p^2} + {b^2} \\\ \Rightarrow {b^2} - {h^2} = - {p^2} \\\$$

Now, we put the ${b^2} - {h^2}$ value in the above equation.
Then,
$\Rightarrow - \dfrac{{{p^2}}}{{{b^2}}}$
We know that $\dfrac{p}{b} = \tan x$. Put that value in above.
Then,
$\therefore - {\tan ^2}x$
Therefore, $1 - {\sec ^2}x$ is simplified as $- {\tan ^2}x$.

Note: If we want to find the result of trigonometric identities, then you are suggested to use the right-angle triangle and Pythagoras theorem. By using these two, we can easily find the result of any trigonometric identity.