Answer:
a) 1
b) [tex]\frac{4}{3}[/tex]
c) = 1
Step-by-step explanation:
We are given the following in the question:
[tex]\cos \theta = \dfrac{6}{8}[/tex]
θ is in the IV quadrant.
[tex]\sin^2 \theta + \cos^2 \theta = 1\\\\\sin \theta = \sqrt{1-\dfrac{36}{64}} = -\dfrac{2\sqrt7}{8}\\\\\tan \theta = \dfrac{\sin \theta}{\cos \theta} = -\dfrac{2\sqrt7}{6}\\\\\csc \theta = \dfrac{1}{\sin \theta} = -\dfrac{8}{2\sqrt7}[/tex]
Evaluate the following:
a)
[tex]\tan \theta\times \cot \theta =\tan \theta\times\dfrac{1}{\tan \theta} = 1[/tex]
b)
[tex]\csc \theta\times \tan \theta\\\\= -\dfrac{8}{2\sqrt7}\times -\dfrac{2\sqrt7}{6} = \dfrac{4}{3}[/tex]
c)
[tex]\sin^2 \theta + \cos^2 \theta = 1\\\text{using the trignometric identity}[/tex]
