given that $\tan\theta = - 1$, what is the value of $sec\theta$, for $\frac{3pi}{2}<\theta<2pi$?\n…

Explanation:

Step1: Recall the identity $\tan\theta=\frac{\sin\theta}{\cos\theta}$

Given $\tan\theta = - 1=\frac{\sin\theta}{\cos\theta}$, so $\sin\theta=-\cos\theta$.

Step2: Use the Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta = 1$

Substitute $\sin\theta = -\cos\theta$ into $\sin^{2}\theta+\cos^{2}\theta = 1$. We get $(-\cos\theta)^{2}+\cos^{2}\theta=1$, which simplifies to $2\cos^{2}\theta = 1$. Then $\cos^{2}\theta=\frac{1}{2}$, and $\cos\theta=\pm\frac{1}{\sqrt{2}}$.

Step3: Determine the sign of $\cos\theta$ based on the range of $\theta$

Since $\frac{3\pi}{2}<\theta<2\pi$, $\theta$ is in the fourth - quadrant. In the fourth - quadrant, $\cos\theta>0$. So $\cos\theta=\frac{1}{\sqrt{2}}$.

Step4: Recall the definition of $\sec\theta$

We know that $\sec\theta=\frac{1}{\cos\theta}$. Substituting $\cos\theta=\frac{1}{\sqrt{2}}$ into the formula, we get $\sec\theta=\sqrt{2}$.

Answer:

B. $\sqrt{2}$