How do you get the exact value of #sec^-1(-2)#?

When working with inverse trig functions, it is better to reverse engineer slightly before you actually evaluate them. In your particular case, this would be rewriting as follows:

#sec(x) = -2#

Keep in mind that you could use any variable for x, I just chose x out of personal preference.

Now, because I've memorised the unit circle, I find it easier to work with sine, cosine and tangent functions. Therefore, I always want to try and get those functions. So, I will rewrite this as:

#1/cos(x) = -2#

Now, if I just go ahead and do some algebra, I get:

#cos(x) = -1/2#

Look familiar? Now we could stop right here and use our unit circle, but since we're talking about inverse trig, I will take it forward just one more step:

#x = cos^-1(-1/2)#

The final answer to this would be #(2pi)/3 + 2pin, (4pi)/3 + 2pin#

If you're unsure how we derived this final answer with the unit circle, or have trouble memorising it, I'd encourage you to watch my video .

Hope that helped :)