To prove that \( 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \) given that \( \tan \theta = \frac{1}{\sqrt{3}} \), follow these steps:
1. Find \(\sin \theta\) and \(\cos \theta\) from \(\tan \theta\):
- We are given \(\tan \theta = \frac{1}{\sqrt{3}}\). Knowing that \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we can set up the relationship:
[tex]\[
\frac{\sin \theta}{\cos \theta} = \frac{1}{\sqrt{3}}
\][/tex]
- This implies:
[tex]\[
\sin \theta = \frac{\cos \theta}{\sqrt{3}}
\][/tex]
2. Use the Pythagorean identity:
- The Pythagorean identity states:
[tex]\[
\sin^2 \theta + \cos^2 \theta = 1
\][/tex]
- Substitute \(\sin \theta = \frac{\cos \theta}{\sqrt{3}}\) into this identity:
[tex]\[
\left(\frac{\cos \theta}{\sqrt{3}}\right)^2 + \cos^2 \theta = 1
\][/tex]
- Simplify the equation:
[tex]\[
\frac{\cos^2 \theta}{3} + \cos^2 \theta = 1
\][/tex]
[tex]\[
\frac{\cos^2 \theta}{3} + \frac{3\cos^2 \theta}{3} = 1
\][/tex]
[tex]\[
\frac{4\cos^2 \theta}{3} = 1
\][/tex]
[tex]\[
\cos^2 \theta = \frac{3}{4}
\][/tex]
3. Determine \(\sin^2 \theta\):
- Substitute \(\cos^2 \theta = \frac{3}{4}\) back into the Pythagorean identity:
[tex]\[
\sin^2 \theta = 1 - \cos^2 \theta
\][/tex]
[tex]\[
\sin^2 \theta = 1 - \frac{3}{4}
\][/tex]
[tex]\[
\sin^2 \theta = \frac{1}{4}
\][/tex]
4. Verify the given equation:
- Substitute \(\sin^2 \theta = \frac{1}{4}\) and \(\cos^2 \theta = \frac{3}{4}\) into the equation \(7 \sin^2 \theta + 3 \cos^2 \theta\):
[tex]\[
7 \sin^2 \theta + 3 \cos^2 \theta = 7 \left(\frac{1}{4}\right) + 3 \left(\frac{3}{4}\right)
\][/tex]
[tex]\[
= 7 \cdot \frac{1}{4} + 3 \cdot \frac{3}{4}
\][/tex]
[tex]\[
= \frac{7}{4} + \frac{9}{4}
\][/tex]
[tex]\[
= \frac{16}{4}
\][/tex]
[tex]\[
= 4
\][/tex]
Thus, we have proved that [tex]\( 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \)[/tex] given that [tex]\( \tan \theta = \frac{1}{\sqrt{3}} \)[/tex].
