Question

Prove cosecant left-parenthesis Start Fraction pi over 2 End Fraction minus x right-parenthesis equals secant x.
(1 point)
Responses

cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction cosine x plus cosine StartFraction pi over 2 EndFraction sine x EndFraction equals secant x
Image with alt text: cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction cosine x plus cosine StartFraction pi over 2 EndFraction sine x EndFraction equals secant x

cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction sine x minus cosine StartFraction pi over 2 EndFraction cosine x EndFraction equals secant x
Image with alt text: cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction sine x minus cosine StartFraction pi over 2 EndFraction cosine x EndFraction equals secant x

cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction sine x plus cosine StartFraction pi over 2 EndFraction cosine x EndFraction equals secant x
Image with alt text: cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction sine x plus cosine StartFraction pi over 2 EndFraction cosine x EndFraction equals secant x

cosecant left-parenthesis StartFraction pi over 2 EndFraction minus x right-parenthesis equals StartFraction 1 over sine StartFraction pi over 2 EndFraction cosine x minus cosine StartFraction pi over 2 EndFraction sine x EndFraction equals secant x

Answer 1

To prove that \(\csc\left(\frac{\pi}{2} - x\right) = \sec x\), we can use some trigonometric identities.

Recall the definitions:

- \(\csc \theta = \frac{1}{\sin \theta}\)
- \(\sec \theta = \frac{1}{\cos \theta}\)

Using the co-function identity for sine:

\[
\sin\left(\frac{\pi}{2} - x\right) = \cos x
\]

Thus, we can express cosecant in terms of sine:

\[
\csc\left(\frac{\pi}{2} - x\right) = \frac{1}{\sin\left(\frac{\pi}{2} - x\right)} = \frac{1}{\cos x}
\]

Next, we know that:

\[
\sec x = \frac{1}{\cos x}
\]

Putting it all together, we find:

\[
\csc\left(\frac{\pi}{2} - x\right) = \frac{1}{\cos x} = \sec x
\]

Thus, we have proven that:

\[
\csc\left(\frac{\pi}{2} - x\right) = \sec x
\]

The correct approach is:

\[
\csc\left(\frac{\pi}{2} - x\right) = \frac{1}{\cos x} = \sec x
\]

So the answer is:

\(\csc\left(\frac{\pi}{2} - x\right) = \sec x\)