Evaluate (if possible) the sine, cosine, and tangent of the real number.

t = - \frac { \pi } { 2 }

t = - \frac { \pi } { 2 }

corresponds to the point

( x , y ) = ( 0,0 )

\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = 0 \\\cos \left( - \frac { \pi } { 2 } \right) = 0 \\\tan \left( - \frac { \pi } { 2 } \right) = 0\end{array}

t = - \frac { \pi } { 2 }

corresponds to the point

( x , y ) = ( - 1,0 )

\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = - 1 \\\cos \left( - \frac { \pi } { 2 } \right) = 0 \\\tan \left( - \frac { \pi } { 2 } \right) = 0\end{array}

t = - \frac { \pi } { 2 }

corresponds to the point

( x , y ) = ( 0 , - 1 )

\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = - 1 \\\cos \left( - \frac { \pi } { 2 } \right) = 0 \\\tan \left( - \frac { \pi } { 2 } \right) \text { is undefined }\end{array}

t = - \frac { \pi } { 2 }

corresponds to the point

( x , y ) = ( - 1,0 )

\begin{array} { l } \sin \left( - \frac { \pi } { 2 } \right) = 0 \\\cos \left( - \frac { \pi } { 2 } \right) = - 1 \\\tan \left( - \frac { \pi } { 2 } \right) = 0\end{array}

E)Not possible

Use the Value of the Trigonometric Function to Find sec⁡(−t)\sec ( - t )sec(−t)

Multiple Choice