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Given [tex]\cos (t) = \frac{1}{2}[/tex] and [tex]\tan(t) \ \textgreater \ 0[/tex], find [tex]\sin (t)[/tex] and [tex]\sec (t)[/tex].

A. [tex]\sin (t) = \frac{\sqrt{3}}{2}, \sec (t) = 2[/tex]

B. [tex]\sin (t) = -\frac{\sqrt{3}}{2}, \sec (t) = -2[/tex]

C. [tex]\sin (t) = \frac{\sqrt{3}}{2}, \sec (t) = -2[/tex]

D. [tex]\sin (t) = -\frac{\sqrt{3}}{2}, \sec (t) = 2[/tex]


Sagot :

To solve the problem of finding [tex]\(\sin(t)\)[/tex] and [tex]\(\sec(t)\)[/tex] given that [tex]\(\cos(t) = \frac{1}{2}\)[/tex] and [tex]\(\tan(t) > 0\)[/tex], we will follow these steps:

1. Determine the quadrant:
- [tex]\(\cos(t) > 0\)[/tex] occurs in the first and fourth quadrants.
- [tex]\(\tan(t) > 0\)[/tex] occurs in the first and third quadrants.

Since we need both conditions to be true simultaneously, the only viable quadrant is the first quadrant where [tex]\(\cos(t) > 0\)[/tex] and [tex]\(\tan(t) > 0\)[/tex].

2. Find [tex]\(\sin(t)\)[/tex]:
- In the first quadrant, all trigonometric functions are positive.
- We use the Pythagorean identity [tex]\(\cos^2(t) + \sin^2(t) = 1\)[/tex].

Given [tex]\(\cos(t) = \frac{1}{2}\)[/tex], we can find [tex]\(\sin(t)\)[/tex] as follows:
[tex]\[ \sin^2(t) = 1 - \cos^2(t) \][/tex]
[tex]\[ \sin^2(t) = 1 - \left(\frac{1}{2}\right)^2 \][/tex]
[tex]\[ \sin^2(t) = 1 - \frac{1}{4} \][/tex]
[tex]\[ \sin^2(t) = \frac{3}{4} \][/tex]
[tex]\[ \sin(t) = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \][/tex]

Since we are in the first quadrant, [tex]\(\sin(t)\)[/tex] is positive. Hence, [tex]\(\sin(t) = \frac{\sqrt{3}}{2}\)[/tex].

3. Find [tex]\(\sec(t)\)[/tex]:
- The secant function is the reciprocal of the cosine function: [tex]\(\sec(t) = \frac{1}{\cos(t)}\)[/tex].

Given [tex]\(\cos(t) = \frac{1}{2}\)[/tex],
[tex]\[ \sec(t) = \frac{1}{\frac{1}{2}} = 2 \][/tex]

Thus, the correct values are:
[tex]\[ \sin(t) = \frac{\sqrt{3}}{2}, \quad \sec(t) = 2 \][/tex]

So, among the given options, the correct one is:
[tex]\[ \sin(t) = \frac{\sqrt{3}}{2}, \sec(t) = 2 \][/tex]
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