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Let theta be an angle in quadrant II such that cos theta = -2/3
Find the exact values of csc theta and tan theta.

Respuesta :

Answer:

So we have [tex]\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}[/tex].

Step-by-step explanation:

Ok so we are in quadrant 2, that means sine is positive while cosine is negative.

We are given [tex]\cos(\theta)=\frac{-2}{3}(\frac{\text{adjacent}}{\text{hypotenuse}})[/tex].

So to find the opposite we will just use the Pythagorean Theorem.

[tex]a^2+b^2=c^2[/tex]

[tex](2)^2+b^2=(3)^2[/tex]

[tex]4+b^2=9[/tex]

[tex]b^2=5[/tex]

[tex]b=\sqrt{5}[/tex]  This is the opposite side.

Now to find [tex]\csc(\theta)[/tex] and [tex]\tan(\theta)[/tex].

[tex]\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}=\frac{3}{\sqrt{5}}[/tex].

Some teachers do not like the radical on bottom so we will rationalize the denominator by multiplying the numerator and denominator by sqrt(5).

So [tex]\csc(\theta)=\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}}=\frac{3 \sqrt{5}}{5}[/tex].

And now [tex]\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}=\frac{\sqrt{5}}{-2}=\frac{-\sqrt{5}}{2}[/tex].

So we have [tex]\csc(\theta)=\frac{3 \sqrt{5}}{5} \text{ and } \tan(\theta)=\frac{-\sqrt{5}}{2}[/tex].

lublana

Answer:

[tex]tan\theta{3}=-\frac{\sqrt5}{2}[/tex]

[tex]cosec\theta=\frac{3}{\sqrt5}[/tex]

Step-by-step explanation:

We are given that [tex]\theta[/tex] be an  angle in quadrant II and [tex]cos\theta=-\frac{2}{3}[/tex]

We have to find the exact values of [tex]cosec\theta[/tex] and [tex]tan\theta[/tex].

[tex]sec\theta=\frac{1}{cos\theta}[/tex]

Then substitute the value of cos theta and we get

[tex]sec\theta=\frac{1}{-\frac{2}{3}}[/tex]

[tex]sec\theta=-\frac{3}{2}[/tex]

Now, [tex]1+tan^2\theta=sec^2\theta[/tex]

[tex]tan^2\theta=sec^2\theta-1[/tex]

Substitute the value of sec theta then we get

[tex]tan^2\theta= (-\frac{3}{2})^2-1[/tex]

[tex]tan^2\theta=\frac{9}{4}-1=\frac{9-4}{4}=\frac{5}{4}[/tex]

[tex]tan\theta=\sqrt{\frac{5}{4}}=-\frac{\sqrt5}{2}[/tex]

Because[tex] tan\theta [/tex] in quadrant II is negative.

[tex]sin^2\theta=1-cos^2\theta[/tex]

[tex]sin^2\theta=1-(\farc{-2}{3})^2[/tex]

[tex]sin^2\theta=1-\frac{4}{9}[/tex]

[tex]sin^2\theta=\frac{9-4}{9}=\frac{5}{9}[/tex]

[tex]sin\theta=\sqrt{\frac{5}{9}}[/tex]

[tex]sin\theta=\frac{\sqrt5}{3}[/tex]

Because in quadrant II [tex]sin\theta[/tex] is positive.

[tex]cosec\theta=\frac{1}{sin\theta}=\frac{1}{\frac{\sqrt5}{3}}[/tex]

[tex]cosec\theta=\frac{3}{\sqrt5}[/tex]

[tex]cosec\theta[/tex] is positive in II quadrant.

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