How to find inverse cot, sec csc?

Question

How would u find inverse cot csc and sec using a TI83+ because it does not have a button for the functions, for sin cos tan u just do inverse function, what do you do for the cot sec csc (i have already tried 1/inverse sin cos or tant, it doesnt work)

Answer 1

You're doing the inversion in the wrong order. The correct formulas are:

arccot x = arctan (1/x)
arcsec x = arccos (1/x)
arccsc x = arcsin (1/x)

This comes from the fact that, e.g. cot x=1/tan x. So let y=cot x, then 1/y=tan x, taking arctan of both sides yields arctan (1/y)=x±πk (the πk is due to the fact that many angles have the same tangent, and the arctan function only returns one of them). The other formulas have similar derivations.

Answer 2

This Site Might Help You.

RE:
How to find inverse cot, sec csc?
How would u find inverse cot csc and sec using a TI83+ because it does not have a button for the functions, for sin cos tan u just do inverse function, what do you do for the cot sec csc (i have already tried 1/inverse sin cos or tant, it doesnt work)

Answer 3

Inverse Cotangent

Answer 4

For the best answers, search on this site https://shorturl.im/axT6p

1) cos[sin^-1 (-4/5)] solve it in terms of cosine. sin^-1 (-4/5)=x Now we have cos x instead of what we originally had. Solve cos x in terms of sin. Remember that sin²x+cos²x=1. Therefore cosx=√(1-sin²x) So now have cosx=√(1-sin²x) but remember that we said x=sin^-1 (-4/5) so therefore cosx=cos[sin^-1 (-4/5)]=√(1-sin²x) =√(1-sin²([sin^-1 (-4/5)])) =√(1-(-4/5)²)=√(1-(16/25)) =√(9/25)=3/5 The rest can be done the same way.