Find the exact values of the three trignometric functions of the angle $\theta$ $( \sin \theta , \cos \theta , \tan \theta )$ shown in the figure. (Use the Pythagorean Theorem to find the third side of the traingle.)

$a=2$
A)$\sin \theta = \sqrt { 2 } , \tan \theta = 1 , \quad \sec \theta = \frac { \sqrt { 2 } } { 2 }$
B)$\sin \theta = 1 , \tan \theta = \sqrt { 2 } , \quad \sec \theta = \frac { \sqrt { 2 } } { 2 }$
C)$\sin \theta = \frac { \sqrt { 2 } } { 2 } , \tan \theta = \sqrt { 2 } , \sec \theta = 1$
D)$\sin \theta = \frac { \sqrt { 2 } } { 2 } , \tan \theta = 1 , \sec \theta = \sqrt { 2 }$
E)$\sin \theta = 1 , \tan \theta = \frac { \sqrt { 2 } } { 2 } , \sec \theta = \sqrt { 2 }$

Question

Find the exact values of the three trignometric functions of the angle $\theta$ $( \sin \theta , \cos \theta , \tan \theta )$ shown in the figure. (Use the Pythagorean Theorem to find the third side of the traingle.) 
 
$a=2$ 
A)$\sin \theta = \sqrt { 2 } , \tan \theta = 1 , \quad \sec \theta = \frac { \sqrt { 2 } } { 2 }$ 
B)$\sin \theta = 1 , \tan \theta = \sqrt { 2 } , \quad \sec \theta = \frac { \sqrt { 2 } } { 2 }$ 
C)$\sin \theta = \frac { \sqrt { 2 } } { 2 } , \tan \theta = \sqrt { 2 } , \sec \theta = 1$ 
D)$\sin \theta = \frac { \sqrt { 2 } } { 2 } , \tan \theta = 1 , \sec \theta = \sqrt { 2 }$ 
E)$\sin \theta = 1 , \tan \theta = \frac { \sqrt { 2 } } { 2 } , \sec \theta = \sqrt { 2 }$

Quizplus · Accepted Answer

Using the Pythagorean theorem, we find the length of the hypotenuse to be $\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$. Thus, $\sin	heta=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}$. Also, $	an	heta=\dfrac{\sin	heta}{\cos	heta}=\dfrac{2/\left(2\sqrt{2}ight)}{2/\left(2\sqrt{2}ight)}=1$. Finally, $\sec	heta=\dfrac{1}{\cos	heta}=\dfrac{2\sqrt{2}}{2}=2^{3/2}$, so the answer is $\boxed{	extbf{(D)}}$.

Find the Exact Values of the Three Trignometric Functions of the Angle