Domain Of Sec X
Domain Of Sec X. 3) the domain of is the set of all real numbers except , being any integer. Hence, the domain of sec x.
Solution verified by toppr here, f(x)=y=sec −1x so, x=secy now the range of secy is domain of sec −1x i.e secy is possible in r−(−1,1) so the range of secy is r−(−1,1). X r ∣ x = (n π + 2 π ) range: Those points are at pi/2 and (3pi)/2.
Sec X Will Not Be Defined At The Points Where Cos X Is 0.
4) the range of is given. Hence, the domain of sec x. Sec x = 1/cos x cosec x = 1/sin x cot x = 1/tan x hence, these ratios will not be defined for the following:
Sec Inverse X Is An Important Inverse Trigonometric Function.
6 rows sec inverse x. Set the argument in sec(sin(x)) sec ( sin ( x)) equal to π 2 +πn π 2 + π n to find where the expression is undefined. But doesn't that exclude valid values like 2pi/2?
1) Sec X Has A Period Equal To.
I'm using multiple resources and both say the domain of sec(x) must be restricted, due to the asymptotes, to x != (n pi)/2. Precalculus find the domain and range f (x)=sec (x) f (x) = sec(x) f ( x) = sec ( x) set the argument in sec(x) sec ( x) equal to π 2 +πn π 2 + π n to find where the expression is. From both the graphs it is evident that the domain will be all real.
So Sec X Can Only Take Values That Are ≤ − 1 ∪ Y ≥ 1 Or Y ∣ ( − ∞ , − 1 ) ∪ ( 1.
Sin(x) = π 2 +πn sin ( x) = π 2 + π n, for any integer n n. Therefore, secx would simply be undefined wherever cosx = 0. Solution verified by toppr here, f(x)=y=sec −1x so, x=secy now the range of secy is domain of sec −1x i.e secy is possible in r−(−1,1) so the range of secy is r−(−1,1).
Find The Domain And Range Y=Sec (2X) Y = Sec(2X) Y = Sec ( 2 X) Set The Argument In Sec(2X) Sec ( 2 X) Equal To Π 2 +Πn Π 2 + Π N To Find Where The Expression Is Undefined.
Cosine function only takes values that are between − 1 and + 1. X = (n π + 2 π ) i.e., domain : Similarly sec (x)=1/cos (x) so its domain will also be all real numbers except { (2n+1)}.
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