Respuesta :

let's recall that negative angles go clockwise, so we have -1740°, how many revolutions is that?  1740 - 360 - 360 - 360 - 360 = 300, meaning, that 360 + 360 + 360 + 360 + 300 = 1740.

So, if we move clockwise and go around 4 times over, and then land on -300°, we'll be landing on the 1st Quadrant as you see in the picture below.

[tex]\bf csc(-1740^o)\implies csc(-300^o)\implies csc\left( \frac{\pi }{3} \right)\implies \cfrac{1}{sin\left( \frac{\pi }{3} \right)} \\\\\\ \cfrac{~~1~~}{\frac{\sqrt{3}}{2}}\implies \cfrac{2}{\sqrt{3}}[/tex]

Ver imagen jdoe0001

Answer:

[tex]\frac{2}{\sqrt{3}}[/tex]

Step-by-step explanation:

Consider [tex]\csc (-1740^{\circ})[/tex]

We know that angle [tex]-\theta[/tex] lies in fourth quadrant and [tex]\csc(-\theta )[/tex] is negative in the fourth quadrant .

So, [tex]\csc (-1740^{\circ})=-\csc(1740^{\circ} )[/tex]

We can write [tex]1740^{\circ}[/tex] as [tex]1740^{\circ}=10\pi-60^{\circ}[/tex]

Therefore,

[tex]\csc (-1740^{\circ})=-\csc(1740^{\circ} )=-\csc\left ( 10\pi-60^{\circ} \right )\\\csc(-\theta )[/tex]

Here, [tex]10\pi-60^{\circ}[/tex] lies in fourth quadrant in which cosec function is negative, so [tex]\csc (-1740^{\circ})=-\csc(1740^{\circ} )=-\csc\left ( 10\pi-60^{\circ} \right )\\\csc(\theta )=\csc \left ( 60^{\circ} \right )=\frac{2}{\sqrt{3}}[/tex]

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