Given [tex]\tan \theta = -\frac{4}{3}[/tex] where [tex]\frac{\pi}{2} \ \textless \ \theta \ \textless \ \pi[/tex], what is the value of [tex]\sec \theta[/tex]?

1. [tex]-\frac{5}{3}[/tex]
2. [tex]\frac{4}{5}[/tex]
3. [tex]-\frac{3}{5}[/tex]
4. [tex]\frac{5}{3}[/tex]



Answer :

To determine the value of [tex]\(\sec \theta\)[/tex] given [tex]\(\tan \theta = -\frac{4}{3}\)[/tex] and [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], let’s follow these steps:

1. Understand the given information:
- [tex]\(\tan \theta = -\frac{4}{3}\)[/tex]
- [tex]\(\theta\)[/tex] is in the second quadrant, where the angle lies between [tex]\(\frac{\pi}{2}\)[/tex] and [tex]\(\pi\)[/tex].

2. Identify the relationship between tangent and secant:
We know the trigonometric identity:
[tex]\[ \sec^2 \theta = 1 + \tan^2 \theta \][/tex]

3. Calculate [tex]\(\sec^2 \theta\)[/tex]:
- Given [tex]\(\tan \theta = -\frac{4}{3}\)[/tex],
[tex]\[ \tan^2 \theta = \left( -\frac{4}{3} \right)^2 = \frac{16}{9} \][/tex]
- Apply the identity:
[tex]\[ \sec^2 \theta = 1 + \frac{16}{9} = \frac{9}{9} + \frac{16}{9} = \frac{25}{9} \][/tex]

4. Determine [tex]\(\sec \theta\)[/tex]:
- Take the square root of both sides to solve for [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \pm \sqrt{\frac{25}{9}} = \pm \frac{5}{3} \][/tex]

5. Consider the quadrant:
Since [tex]\(\theta\)[/tex] is in the second quadrant ([tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex]), and we know that in the second quadrant, the cosine function is negative, [tex]\(\sec \theta\)[/tex] (which is the reciprocal of [tex]\(\cos \theta\)[/tex]) must also be negative.

6. Select the correct value:
[tex]\[ \sec \theta = -\frac{5}{3} \][/tex]

Thus, the correct value of [tex]\(\sec \theta\)[/tex] is [tex]\(\boxed{-\frac{5}{3}}\)[/tex].