Question

Given that tangent theta = negative 1, what is the value of secant theta, for StartFraction 3 pi Over 2 EndFraction less-than theta less-than 2 pi?

Answer 1

Answer:

The answer is $\sqrt{2$

Step-by-step explanation:

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Answer 2

Answer:

$sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}$

Step-by-step explanation:

Start by noticing that the angle $\theta$ is on the 4th quadrant (between $\frac{3\pi}{2}$ and $2\pi$. Recall then that in this quadrant the functions tangent and cosine are positive, while the function sine is negative in value. This is important to remember given the fact that tangent of an angle is defined as the quotient of the sine function at that angle divided by the cosine of the same angle:

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}$

Now, let's use the information that the tangent of the angle in question equals "-1", and understand what that angle could be:

$tan(\theta)=\frac{sin(\theta)}{cos(\theta)}\\-1=\frac{sin(\theta)}{cos(\theta)}\\-cos(\theta)=sin(\theta)$

The particular special angle that satisfies this (the magnitude of sine and cosine the same) in the 4th quadrant, is the angle $\frac{7\pi}{4}$

which renders for the cosine function the value $\frac{\sqrt{2} }{2}$.

Now, since we are asked to find the value of the secant of this angle, we need to remember the expression for the secant function in terms of other trig functions: $sec(\theta)=\frac{1}{cos(\theta)}$

Therefore the value of the secant of this angle would be the reciprocal of the cosine of the angle, that is: $sec(\theta)=\frac{2}{\sqrt{2} }=\sqrt{2}$