The range of the function y=secx-2 is all reals except

-1 1 -3 -2

For a function y = f(x), the range is the set of all the possible values of y.

In the question you wrote:

y = secx - 2

This can be interpreted as:

y = sec(x - 2)

or

y = sec(x) - 2

So let's see each case (these are kinda the same)

If the function is:

y = sec(x - 2)

Firs remember that:

sec(x) = 1/cos(x)

then we can rewrite:

y = 1/cos(x - 2)

notice that the function cos(x) has the range -1 ≤ y ≤ 1

Then for the two extremes we have:

y = 1/1 = 1

y = 1/-1 = -1

Notice that for:

y = 1/cos(x - 2)

y can never be in the range  -1 < x < 1

As the denominator cant be larger, in absolute value, than 1.

Then we can conclude that the range is all reals except the interval:

-1 < y < 1

If instead the function was:

y = sec(x) - 2

y = 1/cos(x) - 2

Then with the same reasoning, the range will be the set of all real values except:

-1 - 2 < y < 1 - 2

-3 < y < -1

answer: 15

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.