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If [tex]\cos \theta=\frac{4}{7}[/tex] and [tex]\csc \theta\ \textless \ 0[/tex], find [tex]\sin \theta[/tex] and [tex]\tan \theta[/tex].

A. [tex]\sin \theta=4, \tan \theta=\frac{4 \sqrt{33}}{7}[/tex]

B. [tex]\csc \theta=4 \sqrt{7}, \tan \theta=\frac{33}{7}[/tex]

C. [tex]\sin \theta=\frac{-\sqrt{33}}{7}, \tan \theta=\frac{-\sqrt{33}}{4}[/tex]

D. [tex]\sin \theta=\frac{4}{7}, \tan \theta=\frac{7}{4}[/tex]



Answer :

GinnyAnswer
To find [tex]\(\sin \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] given that [tex]\(\cos \theta = \frac{4}{7}\)[/tex] and [tex]\(\csc \theta < 0\)[/tex], we can follow these steps carefully:

1. Recognize the given information:
[tex]\[\cos \theta = \frac{4}{7}\][/tex]
[tex]\[\csc \theta < 0 \implies \sin \theta < 0\][/tex]

2. Use the Pythagorean identity to find [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[\sin^2 \theta + \cos^2 \theta = 1\][/tex]
Plugging in the given value for [tex]\(\cos \theta\)[/tex]:
[tex]\[\sin^2 \theta + \left(\frac{4}{7}\right)^2 = 1\][/tex]
Calculate [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[\left(\frac{4}{7}\right)^2 = \frac{16}{49}\][/tex]
Thus:
[tex]\[\sin^2 \theta + \frac{16}{49} = 1\][/tex]

3. Isolate [tex]\(\sin^2 \theta\)[/tex]:
[tex]\[\sin^2 \theta = 1 - \frac{16}{49}\][/tex]
Convert 1 to a fraction with a denominator of 49:
[tex]\[1 = \frac{49}{49}\][/tex]
So:
[tex]\[\sin^2 \theta = \frac{49}{49} - \frac{16}{49} = \frac{33}{49}\][/tex]

4. Solve for [tex]\(\sin \theta\)[/tex]:
[tex]\[\sin \theta = \pm \sqrt{\frac{33}{49}}\][/tex]
Since [tex]\(\sin \theta < 0\)[/tex], we take the negative root:
[tex]\[\sin \theta = -\sqrt{\frac{33}{49}} = -\frac{\sqrt{33}}{7}\][/tex]

5. Now, find [tex]\(\tan \theta\)[/tex] using the definition of tangent:
[tex]\[\tan \theta = \frac{\sin \theta}{\cos \theta}\][/tex]
Substituting the values we have:
[tex]\[\tan \theta = \frac{-\frac{\sqrt{33}}{7}}{\frac{4}{7}} = \frac{-\sqrt{33}}{7} \times \frac{7}{4} = \frac{-\sqrt{33}}{4}\][/tex]

So, the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\tan \theta\)[/tex] are:
[tex]\[ \boxed{\sin \theta = \frac{-\sqrt{33}}{7}, \tan \theta = \frac{-\sqrt{33}}{4}} \][/tex]

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