Ł 5.7 .2 Test (CST): Trigonometry Question 10 of 25 Which function does not have a period of \( 2 \pi \) ? A. \( y=\sec x \) B. \( y=\csc x \) C. \( y=\cos x \) D. \( y=\tan x \)

To determine which function does not have a period of \( 2\pi \), let's analyze the periodicity of each function: 1. **\( y = \sec x \)**: - **Period**: \( 2\pi \) - **Explanation**: Secant is the reciprocal of cosine. Since cosine has a period of \( 2\pi \), secant also shares this period. 2. **\( y = \csc x \)**: - **Period**: \( 2\pi \) - **Explanation**: Cosecant is the reciprocal of sine. Sine has a period of \( 2\pi \), so cosecant inherits the same period. 3. **\( y = \cos x \)**: - **Period**: \( 2\pi \) - **Explanation**: The cosine function naturally has a period of \( 2\pi \). 4. **\( y = \tan x \)**: - **Period**: \( \pi \) - **Explanation**: Tangent is the ratio of sine to cosine. It has a shorter period of \( \pi \) because it repeats its values every \( \pi \) radians, not every \( 2\pi \) radians. **Conclusion**: Among the given options, **\( y = \tan x \)** does not have a period of \( 2\pi \). Instead, its period is \( \pi \). **Answer**: D. \( y=\tan x \)