Question

1. for the point $(5, - 10)$ on the terminal arm of an angle $\theta$ in standard position:

a) determine the exact values of the three reciprocal trigonometric ratios. (2 marks)

Explanation:

Step1: Calculate the value of \(r\)

For a point \((x,y)\) in the coordinate plane, \(r=\sqrt{x^{2}+y^{2}}\). Given \(x = 5\) and \(y=-10\), then \(r=\sqrt{5^{2}+(-10)^{2}}=\sqrt{25 + 100}=\sqrt{125}=5\sqrt{5}\)

Step2: Find the reciprocal trigonometric ratios

• Cosecant (\(\csc\theta\)): \(\csc\theta=\frac{r}{y}\), substituting \(r = 5\sqrt{5}\) and \(y=-10\), we get \(\csc\theta=\frac{5\sqrt{5}}{-10}=-\frac{\sqrt{5}}{2}\)
• Secant (\(\sec\theta\)): \(\sec\theta=\frac{r}{x}\), substituting \(r = 5\sqrt{5}\) and \(x = 5\), we get \(\sec\theta=\frac{5\sqrt{5}}{5}=\sqrt{5}\)
• Cotangent (\(\cot\theta\)): \(\cot\theta=\frac{x}{y}\), substituting \(x = 5\) and \(y=-10\), we get \(\cot\theta=\frac{5}{-10}=-\frac{1}{2}\)

Answer:

\(\csc\theta=-\frac{\sqrt{5}}{2}\), \(\sec\theta=\sqrt{5}\), \(\cot\theta=-\frac{1}{2}\)