If cos \(\theta\) = \(\frac{\text{x}}{\text{y}}\), find tan \(\theta\) in terms of x and y

The question gives the value of \(\cos \theta\) as \(\frac{x}{y}\), and asks for \(\tan \theta\) in terms of \(x\) and \(y\).

Recall the definitions from trigonometry in a right triangle:

• \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
• \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)
• \(\tan \theta = \frac{\text{opposite side}}{\text{adjacent side}}\)

So, if \(\cos \theta = \frac{x}{y}\), then:

• The adjacent side has length \(x\).
• The hypotenuse has length \(y\).

We need the opposite side to find \(\tan \theta\). Use the Pythagorean theorem for a right triangle:

\[ \text{(hypotenuse)}^2 = (\text{adjacent})^2 + (\text{opposite})^2 \]

Plug in the values:

\[ y^2 = x^2 + (\text{opposite})^2 \]

Solve for the opposite side:

\[ (\text{opposite})^2 = y^2 - x^2 \] \[ \text{opposite} = \sqrt{y^2 - x^2} \]

Now, put this into the ratio for \(\tan \theta\):

\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{\sqrt{y^2 - x^2}}{x} \]

This formula comes from the relationships among the sides in a right triangle. It uses the fact that the square of the hypotenuse minus the square of the adjacent side gives the square of the opposite side, and then takes the square root to get its length. That's why \(\tan \theta\) in terms of \(x\) and \(y\) is \(\frac{\sqrt{y^2 - x^2}}{x}\).