Question

Evaluating the Six Trigonometric Function Question attached

Answer 1

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.


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


`\bf sec\left( (7\pi )/(4) \right)\implies \cfrac{1}{cos\left( (7\pi )/(4) \right)}\implies \cfrac{1}{(√(2))/(2)}\implies \cfrac{(1)/(1)}{(√(2))/(2)}\implies \cfrac{1}{1}\cdot \cfrac{2}{√(2)}\implies \cfrac{2}{√(2)} \\\\\\ \stackrel{\textit{rationalizing the denominator}}{\cfrac{2}{√(2)}\cdot \cfrac{√(2)}{√(2)}\implies \cfrac{2√(2)}{(√(2))^2}\implies \cfrac{2√(2)}{2}\implies √(2)}`

Answer 2

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)