# How do you find #cot(sin^-1(p/q))#?

##### 4 Answers

#### Explanation:

Here,

Let

Q4#, when p and q have opposite signs.

Here, in both quadrants, cosine is positive. So, cotangent and sine

have the same sign.

The given expression is

In brief, the negative sign is necessary, despite p > 0.

For example, if p = 1 and

#### Explanation:

Let's start with the original:

First thing to do is to understand what

The question is asking for the cotangent, which is Adjacent side

We can find the Adjacent in terms of the Opposite and Hypotenuse using the Pythagorean Theorem:

We now know all the sides so we can answer the cotangent part of the question:

If we are told that

#sqrt(q^2-p^2)/p#

#### Explanation:

If

where

#sin theta = "opposite"/"hypotenuse" = p/q#

#cot theta = "adjacent"/"opposite" = sqrt(q^2-p^2)/p#

If we cannot assume

Given Real number

If

#cos theta = sqrt(1 - sin^2 theta) = sqrt(1 - s^2)#

So we find:

#cot(sin^(-1) s) = cot theta = cos theta / sin theta = sqrt(1-s^2)/s#

Substituting

#cot(sin^(-1) (p/q)) = sqrt(1-p^2/q^2)/(p/q)#

This will have the correct sign, regardless of whether

If we are additionally told that

#cot(sin^(-1) (p/q)) = sqrt(1-p^2/q^2)/(p/q)#

#color(white)(cot(sin^(-1) (p/q))) = sqrt(1-p^2/q^2)/(p/q) * q/q#

#color(white)(cot(sin^(-1) (p/q))) = sqrt(q^2-p^2)/p#