Question

9Let O be an angle in quadrant II such that csc =4"Find the exact values of tan 0 and cos 0.

Answer 1

Answer: cscθ = 4/√15, cot=-1/√15.

Explanation:

For this question we will use the following diagram:

Using the Pythagorean theorem we know that:

$l^2+(-\frac{1}{4})^2=1^2.$

Solving for l we get:

$\begin{gathered} l^2=1-\frac{1}{16}=\frac{15}{16}, \\ l=\frac{\sqrt[]{15}}{4}. \end{gathered}$

Therefore:

$\csc \theta=\frac{1}{l}=\frac{1}{\frac{\sqrt[]{15}}{4}}=\frac{4}{\sqrt[]{15}}.$$\cot \theta=-\frac{\frac{1}{4}}{l}=-\frac{4}{4\sqrt[]{15}}=-\frac{1}{\sqrt[]{15}}.$