These are the last questions i need to finish evaluating trigonometric equation?

Question

I need some help.
just answer the ones that you can do.
1.) tan^-1(-1)
2.) sec^2(11pi/6)
3.) sin(csc^-1(-2))
4.) sec^-1(sec(-30º))

Answer 1

1) tan^(-1)(-1).

I'm going to assume you mean the inverse of tan, which is also known as arctan.

arctan(-1)

By defintion, y = arctan(x) if and only if
tan(y) = x, for -&pi/2 < x < &pi/2. Therefore, if

tan(y) = -1, then we want a solution in quadrants 4 and 1. This is only true at the point y = -&pi/4 (quadrant 4), and remember that it lies in the given interval. 7&pi/4 also satisfies the above solution, but falls outside of the interval -&pi/2 < x < &pi/2, so we reject that solution.

2) sec^2(11&pi/6)

This is the same as

[ sec(11&pi/6) ]^2

Secant is the reciprocal of cosine, so this is equal to

[ 1/cos(11&pi/6) ]^2

And the cosine of 11&pi/6 is in the 4th quadrant, and equal to sqrt(3)/2, so we get

[ 1/[ sqrt(3)/2 ] ]^2

And 1 over anything is its reciprocal.

[ 2/sqrt(3) ]^2

And which we now square.

4/3

3) sin( csc^(-1)(-2) )

To solve this, use trig SOHCAHTOA in this way.

Let t = csc^(-1)(-2). Then
csc(t) = -2. Taking the reciprocal of both sides,

sin(t) = -1/2

And as you can see,

sin(t) = sin( csc^(-1)(-2) ) = -1/2

Answer 2

tan^-1(-1) = -p/4+kpi
2)1/cos^2(11pi/6)= 1/cos^2(pi/6) = 4/3
3) cosec = 1/sin =-2 so sin = -1/2 and the angles are
5pi/4 +2kpi and 7pi/4+2kpi and sin =-1/2
sec(-30º=1/cos(30) = 2sqrt3/3
angle=-30+2kpi and 30+2kpi