Answer :
To demonstrate that [tex]\(\sqrt{q^2 - p^2} \tan \theta = p\)[/tex] given [tex]\(\sin \theta = \frac{p}{q}\)[/tex], we need to follow a series of trigonometric identity manipulations. Here's the step-by-step solution:
1. Express [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Recall that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
2. Given: [tex]\(\sin \theta = \frac{p}{q}\)[/tex].
3. Find [tex]\(\cos \theta\)[/tex]:
We use the Pythagorean identity that relates [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substituting [tex]\(\sin \theta = \frac{p}{q}\)[/tex]:
[tex]\[ \left(\frac{p}{q}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{p^2}{q^2} + \cos^2 \theta = 1 \][/tex]
Solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta = 1 - \frac{p^2}{q^2} \][/tex]
[tex]\[ \cos^2 \theta = \frac{q^2 - p^2}{q^2} \][/tex]
Take the square root of both sides:
[tex]\[ \cos \theta = \sqrt{\frac{q^2 - p^2}{q^2}} = \frac{\sqrt{q^2 - p^2}}{q} \][/tex]
4. Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] into [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{p}{q}}{\frac{\sqrt{q^2 - p^2}}{q}} = \frac{p}{\sqrt{q^2 - p^2}} \][/tex]
5. Multiply [tex]\(\tan \theta\)[/tex] by [tex]\(\sqrt{q^2 - p^2}\)[/tex]:
[tex]\[ \sqrt{q^2 - p^2} \cdot \tan \theta = \sqrt{q^2 - p^2} \cdot \frac{p}{\sqrt{q^2 - p^2}} \][/tex]
The [tex]\(\sqrt{q^2 - p^2}\)[/tex] terms cancel each other out:
[tex]\[ \sqrt{q^2 - p^2} \cdot \frac{p}{\sqrt{q^2 - p^2}} = p \][/tex]
Therefore, we have shown that:
[tex]\[ \sqrt{q^2 - p^2} \tan \theta = p \][/tex]
The expression is indeed proved.
1. Express [tex]\(\tan \theta\)[/tex] in terms of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Recall that [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta}\)[/tex].
2. Given: [tex]\(\sin \theta = \frac{p}{q}\)[/tex].
3. Find [tex]\(\cos \theta\)[/tex]:
We use the Pythagorean identity that relates [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
[tex]\[ \sin^2 \theta + \cos^2 \theta = 1 \][/tex]
Substituting [tex]\(\sin \theta = \frac{p}{q}\)[/tex]:
[tex]\[ \left(\frac{p}{q}\right)^2 + \cos^2 \theta = 1 \][/tex]
[tex]\[ \frac{p^2}{q^2} + \cos^2 \theta = 1 \][/tex]
Solve for [tex]\(\cos^2 \theta\)[/tex]:
[tex]\[ \cos^2 \theta = 1 - \frac{p^2}{q^2} \][/tex]
[tex]\[ \cos^2 \theta = \frac{q^2 - p^2}{q^2} \][/tex]
Take the square root of both sides:
[tex]\[ \cos \theta = \sqrt{\frac{q^2 - p^2}{q^2}} = \frac{\sqrt{q^2 - p^2}}{q} \][/tex]
4. Substitute [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] into [tex]\(\tan \theta\)[/tex]:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{p}{q}}{\frac{\sqrt{q^2 - p^2}}{q}} = \frac{p}{\sqrt{q^2 - p^2}} \][/tex]
5. Multiply [tex]\(\tan \theta\)[/tex] by [tex]\(\sqrt{q^2 - p^2}\)[/tex]:
[tex]\[ \sqrt{q^2 - p^2} \cdot \tan \theta = \sqrt{q^2 - p^2} \cdot \frac{p}{\sqrt{q^2 - p^2}} \][/tex]
The [tex]\(\sqrt{q^2 - p^2}\)[/tex] terms cancel each other out:
[tex]\[ \sqrt{q^2 - p^2} \cdot \frac{p}{\sqrt{q^2 - p^2}} = p \][/tex]
Therefore, we have shown that:
[tex]\[ \sqrt{q^2 - p^2} \tan \theta = p \][/tex]
The expression is indeed proved.
