Using the unit circle, how do you find the value of the trigonometric function: $sec - (\frac{π}{2})$ $sec -(pi/2)$ ?

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Using the unit circle, how do you find the value of the trigonometric function: $sec - (\frac{π}{2})$ $sec -(pi/2)$ ?

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Answer 1 · 2018-03-12T02:54:45

$sec - (\frac{π}{2}) = \infty$ $sec -(pi/2) = oo$

Explanation:

enter image source here

$- (\frac{π}{2}) = 2 π - \frac{π}{2} = \frac{3 π}{2}$ $-(pi/2) = 2pi - pi/2 = (3pi)/2$ . It’s in III quadrant.

As per unit circle, coordinates are (0,-1)#

#cos theta = enter image source here

$cos (- \frac{π}{2}) = cos (\frac{π}{2}) = 0$ $cos (-pi/2) = cos (pi/2) = 0$

Hence $sec - (\frac{π}{2}) = \frac{1}{cos - (\frac{π}{2})} = \frac{1}{cos (\frac{π}{2})} = \frac{1}{0} = \infty$ $sec - (pi/2) = 1 / (cos -(pi/2)) = 1 / cos (pi/2) = 1/ 0 = oo$

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Using the unit circle, how do you find the value of the trigonometric function: $sec - (\frac{π}{2})$ $sec -(pi/2)$ ?

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Explanation:

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