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Graph y = sec ( x )
Solution: As with tangent we will have to avoid x's for which cosine is zero (recall that sec x =1/ cos x)
Secant will not present at
x = ........, - 5 ∏/2 , - 3 ∏/2 , - ∏/2 , ∏/2, 3 ∏/2 , 5 ∏/2 ,........
and the graph will have asymptotes at these points. Following is the graph of secant on the range - 5 ∏/2 Notice as well that the graph is always greater than 1 & less than -1. It should not be terribly surprising. Remember that -1 ≤ cos ( x ) ≤ 1 . Hence, one divided by something less than one will be greater than 1. Also, 1 / ±1 = ±1 and hence we get the following ranges for secant. sec (? x) ≥ 1 and sec (? x) ≤ -1
Notice as well that the graph is always greater than 1 & less than -1. It should not be terribly surprising. Remember that -1 ≤ cos ( x ) ≤ 1 . Hence, one divided by something less than one will be greater than 1. Also, 1 / ±1 = ±1 and hence we get the following ranges for secant.
sec (? x) ≥ 1 and sec (? x) ≤ -1
were can get this please
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7 divided by 66.5
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