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If tan theta is -1, we know immediately that theta is in either Quadrant II or Q IV.  We need to focus on Q IV due to the restrictions on theta.

Because tan theta is -1, the ray representing theta makes a 45 degree angle with the horiz axis, and a 45 degree angle with the negative vert. axis.  Thus the hypotenuse, by the Pythagorean Theorem, tells us that the hyp is sqrt(2).
Thus, the cosine of theta is adj / hyp, or +1 / sqrt(2), or [sqrt(2)]/2

The secant of theta is the reciprocal of that, and thus is

   2            sqrt(2)
---------- * ------------ = sqrt(2)    (answer)
 sqrt(2)      sqrt(2)  
3π/2 < θ < 2π    is another way to say θ is in the IV quadrant.

now, the only time the tangent is 1 or -1, is right in the middle of the quadrant, in this case, that'd be at θ = 7π/4.

[tex]\bf tan\left( \frac{7\pi }{4} \right)=\cfrac{\qquad \stackrel{sine}{\frac{-\sqrt{2}}{2}}\qquad }{\stackrel{cosine}{\frac{\sqrt{2}}{2}}}\implies -1 \\\\\\ \textit{now, let's recall that }sec(\theta )=\cfrac{1}{cos(\theta )}\qquad therefore[/tex]

[tex]\bf sec\left( \frac{7\pi }{4} \right)\implies \cfrac{1}{cos\left( \frac{7\pi }{4} \right)}\implies \cfrac{1}{\frac{\sqrt{2}}{2}}\implies \cfrac{\frac{1}{1}}{\frac{\sqrt{2}}{2}}\implies \cfrac{1}{1}\cdot \cfrac{2}{\sqrt{2}}\implies \cfrac{2}{\sqrt{2}} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{\cfrac{2}{\sqrt{2}}\cdot \cfrac{\sqrt{2}}{\sqrt{2}}\implies \cfrac{2\sqrt{2}}{(\sqrt{2})^2}\implies \cfrac{2\sqrt{2}}{2}\implies \sqrt{2}} [/tex]

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