Question

Suppose that $\theta$ is in standard position and the given point is on the terminal side of $\theta$. Give the exact value of the indicated trig function for $\theta$. (9, 12); Find csc $\theta$. A. $\frac{5}{4}$ B. $\frac{4}{3}$ C. $\frac{5}{3}$ D. $\frac{3}{4}$

          Suppose that $\theta$ is in standard position and the given point is on the terminal side of $\theta$. Give the exact value of the indicated trig function for $\theta$.
(9, 12); Find csc $\theta$.
A. $\frac{5}{4}$
B. $\frac{4}{3}$
C. $\frac{5}{3}$
D. $\frac{3}{4}$
        
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Suppose that θ is in standard position and the given point is on the terminal side of θ. Give the exact value of the indicated trig function for θ.
(9, 12); Find csc θ.
A. (5)/(4)
B. (4)/(3)
C. (5)/(3)
D. (3)/(4)

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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Suppose that θ is in standard position and the given point is on the terminal side of θ. Give the exact value of the indicated trig function for θ. (9,12); Find cscθ A. (5)/(4) B. (4)/(3) C. (5)/(3) D. (3)/(4) Suppose that θ is in standard position and the given point is on the terminal side of θ. Give the exact value of the indicated trig function for (9,12); Find cscθ A. 4 B. 13 C. 3 D. 1
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Transcript

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00:01 Hi, the first we use the relation here, that is cost theta is equal to 1 over 6 theta, right? so we're given here, sec theta is 5 over 3, sec theta is 5 over 3, and it is in 4th quadrant, right? so where we have cause is positive, sec is positive, right? so cause theta is coming out to be 1 over 5 over 3, that is given as 3 over 5, right? so we have here now, we get here, cause theta is coming out to be 3 over 5, and that is coming have to be in the fourth quadrant, right? the next part is cosec theta.
00:36 If you go for the relation here, that is sine square theta plus cose square theta, that equals 1, right? so we'll have sine square theta plus core square theta will be 9 over 25, that equals 1.
00:52 From here we have sine square theta, that equals 1 minus 9 over 25, right? that is given as 25 minus 9.
01:01 60 number 25 this we have so solve here sign theta is coming out to be equal to plus minus 4 over 5 but theta is in fourth quadrant in fourth quadrant sign is negative so we take it as sine theta will be minus 4 over 5 right so option c is not correct right we can eliminate that option a cost it is 3 over 5 yeah we have cos it at 3 over 5 so option is correct right if you look at even cosec -t -t -a, let's find out here cos -t -t -a basically, cosx -t -t -a will be 1 over sine theta, that is given as minus 5 over 4.
01:46 So, option b is correct here...
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