Question

What is a simplified form of the expression Start Fraction secant squared x minus 1 over sine x secant x End Fraction?
(1 point)
Responses

cot x
cot x

csc x
csc x

tan x
tan x

sec x tan x

Answer 1

To simplify the expression \(\frac{\sec^2 x - 1}{\sin x \sec x}\), we can start by using the trigonometric identity \(\sec^2 x - 1 = \tan^2 x\). This gives us:

\[
\sec^2 x - 1 = \tan^2 x
\]

Now, we substitute this into the expression:

\[
\frac{\sec^2 x - 1}{\sin x \sec x} = \frac{\tan^2 x}{\sin x \sec x}
\]

Next, we can express \(\tan x\) and \(\sec x\) in terms of sine and cosine:

\[
\tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}
\]

Now, we substitute these into our expression:

\[
\frac{\tan^2 x}{\sin x \sec x} = \frac{\left(\frac{\sin x}{\cos x}\right)^2}{\sin x \cdot \frac{1}{\cos x}} = \frac{\frac{\sin^2 x}{\cos^2 x}}{\frac{\sin x}{\cos x}}
\]

This can be simplified further:

\[
= \frac{\sin^2 x}{\cos^2 x} \cdot \frac{\cos x}{\sin x} = \frac{\sin x}{\cos^2 x}
\]

Since \(\frac{1}{\cos^2 x} = \sec^2 x\) and also represents the form \(\tan x\):

So, this simplifies to:

\[
\frac{\sin x}{\cos^2 x} = \tan x \sec x
\]

Since we don't have the option for \(\tan x \sec x\), we explore further and verify the main simplification steps.

Returning to \(\sec^2 x\) and rewriting,

\[
\sec x \tan x = \sec x \cdot \frac{\sin x}{\cos x} = \frac{\sin x}{\cos^2 x}
\]

So, the simplified form of the given expression is indeed:

\[
\sec x \tan x
\]

Thus, the correct answer is **sec x tan x**.