Question

If cos(θ) = 6/8 and θ is in the IV quadrant, then fine:

Answer 1

Answer:

a) 1

b) $\frac{4}{3}$

c) = 1

Step-by-step explanation:

We are given the following in the question:

$\cos \theta = \dfrac{6}{8}$

θ is in the IV quadrant.

$\sin^2 \theta + \cos^2 \theta = 1\\\\\sin \theta = \sqrt{1-\dfrac{36}{64}} = -\dfrac{2\sqrt7}{8}\\\\\tan \theta = \dfrac{\sin \theta}{\cos \theta} = -\dfrac{2\sqrt7}{6}\\\\\csc \theta = \dfrac{1}{\sin \theta} = -\dfrac{8}{2\sqrt7}$

Evaluate the following:

a)

$\tan \theta\times \cot \theta =\tan \theta\times\dfrac{1}{\tan \theta} = 1$

b)

$\csc \theta\times \tan \theta\\\\= -\dfrac{8}{2\sqrt7}\times -\dfrac{2\sqrt7}{6} = \dfrac{4}{3}$

c)

$\sin^2 \theta + \cos^2 \theta = 1\\\text{using the trignometric identity}$