Daario
contestada

Find the exact value of csc theta if tan theta = sqrt3 and the terminal side of theta is in Quadrant III.

Respuesta :

LammettHash

For angles [tex]\theta[/tex] in quadrant III, we expect to have [tex]\cos\theta<0[/tex], so we also would have [tex]\sec\theta<0[/tex]. Recall that

[tex]\sin^2\theta+\cos^2\theta=1\implies\tan^2\theta+1=\sec^2\theta[/tex]

[tex]\implies\sec\theta=-\sqrt{\tan^2\theta+1}=-\sqrt{(\sqrt3)^2+1}=-2[/tex]

[tex]\implies\cos\theta=-\dfrac12[/tex]

Then by definition of tangent,

[tex]\tan\theta=\dfrac{\sin\theta}{\cos\theta}\implies\sqrt3=\dfrac{\sin\theta}{-\frac12}\implies\sin\theta=-\dfrac{\sqrt3}2\implies\csc\theta=-\dfrac2{\sqrt3}[/tex]

Answer:

The exact value of    [tex]\csc \theta=-\frac{2}{\sqrt{3}}$[/tex]

Step-by-step explanation:

Trigonometric identities

Trigonometric Identities are the equalities that involve trigonometry functions and holds true for all the values of variables given in the equation. There are various distinct trigonometric identities involving the side length as well as the angle of a triangle.

Given-

[tex]$\sin ^{2} \theta+\cos ^{2} \theta=1[/tex]

[tex]\tan ^{2} \theta+1=\sec ^{2} \theta$[/tex]

[tex]$\ \sec \theta=-\sqrt{\tan ^{2} \theta+1}[/tex]

[tex]=-\sqrt{(\sqrt{3})^{2}+1}=-2$[/tex]

[tex]$\Longrightarrow \cos \theta=-\frac{1}{2}$[/tex]

Then by the trigonometric definition of tanθ .

[tex]$\tan \theta=\frac{\sin \theta}{\cos \theta}[/tex]

[tex]\Longrightarrow \sqrt{3}=\frac{\sin \theta}{-\frac{1}{2}}[/tex]

[tex]\Longrightarrow \sin \theta=-\frac{\sqrt{3}}{2}[/tex]

[tex]\Longrightarrow \csc \theta=-\frac{2}{\sqrt{3}}$[/tex]

Learn more about the Trigonometric identities here-

https://brainly.com/question/20094605

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