Let's start by breaking down the given question and solving it step by step.
Given:
[tex]\[\sin \theta = \frac{p}{q}\][/tex]
We need to show that:
[tex]\[\sqrt{q^2 - p^2} \tan \theta = p\][/tex]
To do this, we will use trigonometric identities.
1. Find [tex]\(\cos \theta\)[/tex] using the Pythagorean identity.
We know that:
[tex]\[\sin^2 \theta + \cos^2 \theta = 1\][/tex]
Given [tex]\(\sin \theta = \frac{p}{q}\)[/tex], we can substitute this into the equation:
[tex]\[\left(\frac{p}{q}\right)^2 + \cos^2 \theta = 1\][/tex]
Simplifying the equation:
[tex]\[\frac{p^2}{q^2} + \cos^2 \theta = 1\][/tex]
Subtract [tex]\(\frac{p^2}{q^2}\)[/tex] from both sides:
[tex]\[\cos^2 \theta = 1 - \frac{p^2}{q^2}\][/tex]
Simplify the right-hand side:
[tex]\[\cos^2 \theta = \frac{q^2 - p^2}{q^2}\][/tex]
Take the square root of both sides to solve for [tex]\(\cos \theta\)[/tex]:
[tex]\[\cos \theta = \sqrt{\frac{q^2 - p^2}{q^2}}\][/tex]
Simplify further:
[tex]\[\cos \theta = \frac{\sqrt{q^2 - p^2}}{q}\][/tex]
2. Find [tex]\(\tan \theta\)[/tex].
The tangent function is defined as:
[tex]\[\tan \theta = \frac{\sin \theta}{\cos \theta}\][/tex]
Substitute the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[\tan \theta = \frac{\frac{p}{q}}{\frac{\sqrt{q^2 - p^2}}{q}}\][/tex]
Simplifying the fraction:
[tex]\[\tan \theta = \frac{p}{q} \cdot \frac{q}{\sqrt{q^2 - p^2}}\][/tex]
[tex]\[\tan \theta = \frac{p}{\sqrt{q^2 - p^2}}\][/tex]
3. Multiply both sides by [tex]\(\sqrt{q^2 - p^2}\)[/tex].
Now, to demonstrate the given expression:
[tex]\[\sqrt{q^2 - p^2} \tan \theta = p\][/tex]
Substitute the value of [tex]\(\tan \theta\)[/tex] we found:
[tex]\[\sqrt{q^2 - p^2} \left(\frac{p}{\sqrt{q^2 - p^2}}\right) = p\][/tex]
Simplify the right-hand side:
[tex]\[\frac{\sqrt{q^2 - p^2} \cdot p}{\sqrt{q^2 - p^2}} = p\][/tex]
[tex]\[p = p\][/tex]
This shows that:
[tex]\[\sqrt{q^2 - p^2} \tan \theta = p\][/tex]
Hence, the required proof is complete.
