Question

Given that tan 0=-1, what is the value of sec 0, for 3pi/2 < 0 < 2pi?

Answer 1

Answer:

√2

Step-by-step explanation:

Given tanθ = -1

θ = arctan(-1)

θ = -45°

Since tan θ is negative in the 2nd and 4th quadrant,

New angles:

θ = 90-45 = 45°(2nd quadrant)

θ = 360-45 = 315° (4the quadrant)

To get secθ between 3π/2< θ<2π

In trigonometry identity,

secθ = 1/cosθ

We will use θ = 315° since it is the only angle that falls within the given range

Sec315° = 1/cos315°

Sec315° = 1/(1/√2)

Sec 315° = √2

Answer 2

Answer:

$\sqrt{2}$

Step-by-step explanation:

Just adjusting the question, since tan (0)=0 and we have a description of a trigonometric interval:

$tan (\theta)=-1, sec(\theta)=?\:for \:\frac{3\pi}{2} < \theta < 2\pi?$

Therefore, let's go for the secant of $\theta$

1). Well, firstly this interval: $\frac{3\pi}{2} < \theta < 2\pi$ is the IV quadrant, where the tangent assumes negative values.

2) One of the notable arcs we have is the

$tan(\frac{\pi}{4})=1$

3) Then If we subtract 360º-45º=315º or $2\pi -\frac{\pi}{4}=\frac{7\pi}{4}$ rad

So this is the arc we want

So we have

In Radians:

$tan (\frac{7\pi}{4})= -1$

In degrees:

$tan(315\º)=-1$

4) Finally, rationalizing radicals on the denominator:

$sec(\theta)=\frac{1}{cos(\theta)}=sec(\frac{7\pi}{4})=\frac{1}{cos(\frac{7\pi}{4})}=\frac{1}{\frac{\sqrt{2}}{2}}\\\\sec(\frac{7\pi}{4})=\frac{1}{\frac{\sqrt{2}}{2}}=1\times \frac{2}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}$