allery$\csc\theta$ (cosecant) is the reciprocal of:
- $\sec\theta$
- $\cos\theta$
- $\tan\theta$
- $\sin\theta$
Show answer and solution
Correct answer: D. $\sin\theta$
$\csc\theta = \dfrac{1}{\sin\theta}$ wherever defined.
NDA Unit 1A1: Relations, Functions and Graphs — practice questions
Explore 5 free Unit 1A1: Relations, Functions and Graphs questions in NDA Mathematics. Open each question or exercise to read its content, reference answers and available explanation.
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$\csc\theta$ (cosecant) is the reciprocal of:Read the answer and worked solution →$\sec\theta \cdot \cos\theta$ equals (where defined):Read the answer and worked solution →Express $40^{\circ} 20'$ in degrees as a fraction.Read the answer and worked solution →$\dfrac{\pi}{6}$ radian = ?Read the answer and worked solution →$\dfrac{3\pi}{2}$ radian = ?Read the answer and worked solution →
Worked practice questions
$\sec\theta \cdot \cos\theta$ equals (where defined):
- $0$
- $1$
- $\sin\theta$
- $\tan\theta$
Show answer and solution
Correct answer: B. $1$
$\sec\theta = 1/\cos\theta$, so $\sec\theta \cdot \cos\theta = (1/\cos\theta) \cdot \cos\theta = 1$.
Express $40^{\circ} 20'$ in degrees as a fraction.
- $\dfrac{121}{3}^{\circ}$
- $40.2^{\circ}$
- $40\dfrac{1}{2}^{\circ}$
- $42^{\circ}$
Show answer and solution
Correct answer: A. $\dfrac{121}{3}^{\circ}$
$20' = \dfrac{20}{60}^{\circ} = \dfrac{1}{3}^{\circ}$.
**So $40^{\circ} 20' = 40 + \dfrac{1}{3} = \dfrac{121}{3}^{\circ}$.**
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