Can some one help me to solve this trig question

1- Indicate the period and the range of the given function.
y = -3 sec x

I know for the period is 2 pi but how can I find the range?

2- Determine the period and equation of asymptote of the function.
y = cot 9pi/4

Can you plz show me the steps? Thanks.

Range of y = sec(x) is (-∞, -1] U [1, ∞)

Therefore, range of y = -3 sec(x) is (-∞, -3] U [3, ∞)

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Are you missing variable "x" in second question? y = cot(9π/4) is constant

I'll assume you mean y = cot(9πx/4)

cot(x) has period = π, so
cot(9πx/4) has period = π / (9π/4) = 4/9

cot(x) has asymptotes when x = πk, for all integers k
cot(9πx/4) has asymptotes when 9πx/4 = πk -----> x = 4/9 k, for all integers k

Mαthmφm

y = -3 sec , the range is
y = cot 9pi/4x , the range is