Evaluate the expressions using reference angles. $\sec \left( \frac { 7 \pi } { 4 } \right)$ $\tan \left( \frac { 7 \pi } { 4 } \right)$
A) $\sec \left( \frac { 7 \pi } { 4 } \right) = \sqrt { 3 }$ $\tan \left( \frac { 7 \pi } { 4 } \right) = 2$
B) $\sec \left( \frac { 7 \pi } { 4 } \right) = \sqrt { 2 }$ $\tan \left( \frac { 7 \pi } { 4 } \right) = - 1$
C) $\sec \left( \frac { 7 \pi } { 4 } \right) = \sqrt { 2 }$ $\tan \left( \frac { 7 \pi } { 4 } \right) = - \frac { 1 } { 2 }$
D) $\sec \left( \frac { 7 \pi } { 4 } \right) = \frac { \sqrt { 2 } } { 2 }$ $\tan \left( \frac { 7 \pi } { 4 } \right) = - 4$
E) $\sec \left( \frac { 7 \pi } { 4 } \right) = \sqrt { 5 }$ $\tan \left( \frac { 7 \pi } { 4 } \right) = - 1$

Question

Quizplus · Accepted Answer

To find the reference angle we need to subtract the angle from the nearest multiple of π or 180 degrees. The nearest multiple of π to 7π/4 is 2π which is equal to 8π/4. So,
Reference angle = (8π/4) - (7π/4) = π/4
Secant of π/4 is equal to √2 and tangent of π/4 is equal to 1. However, we need to consider the quadrant in which the angle lies. As π/4 is in the second quadrant, both secant and tangent are negative. Therefore, 
sec(7π/4) = -√2
tan(7π/4) = -1

Evaluate the Expressions Using Reference Angles sec⁡(7π4)\sec \left( \frac { 7 \pi } { 4 } \right)sec(47π​)

Evaluate the Expressions Using Reference Angles $\sec \left( \frac { 7 \pi } { 4 } \right)$