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:
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:
Now, if I just go ahead and do some algebra, I get:
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:
The final answer to this would be
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 :)