6. First, we need to draw the right triangle. Given that:
[tex]\cos \theta=\frac{2}{\sqrt[]{a+4}}[/tex]
then, the adjacent leg to angle θ measures 2 units, and the hypotenuse measures √(a+4) units. Therefore, the right triangle used as reference is:
Applying the Pythagorean theorem to triangle ABC and solving for leg BC, we get:
[tex]\begin{gathered} AB^2=AC^2+BC^2 \\ \sqrt[]{a+4}^2=2^2+BC^2 \\ a+4=4+BC^2 \\ a+4-4=BC^2 \\ a=BC^2 \\ \sqrt[]{a}=BC \end{gathered}[/tex]
Despite the value of variable a, from the first equation, the cosine of θ is positive. We also know that the sine of θ is positive. With the help of the next table, we can deduce that θ is in the first quadrant
From the table, all the trigonometric functions will be positive.
Using the reference triangle ABC, the values of the trigonometric functions are:
[tex]\begin{gathered} \sin \theta=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{\sqrt[]{a}}{\sqrt[]{a+4}}=\sqrt[]{\frac{a}{a+4}} \\ \csc \theta=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{\sqrt[]{a+4}}{\sqrt[]{a}}=\sqrt[]{\frac{a+4}{a}} \\ \sec \theta=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{\sqrt[]{a+4}}{2} \\ \tan \theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt[]{a}}{2} \\ \cot \theta=\frac{adjacent}{opposite}=\frac{2}{\sqrt[]{a}} \end{gathered}[/tex]
