If ` tan theta = (1)/( sqrt(3))`, prove that ` 7 sin^(2) theta + 3 cos ^(2) theta = 4 `

Answer

To prove that \( 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \) given that \( \tan \theta = \frac{1}{\sqrt{3}} \), we can follow these steps:

### Step 1: Understand the given information
We know that:
\[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{1}{\sqrt{3}} \]
This implies that if we let the opposite side (perpendicular) be \( 1 \) and the adjacent side (base) be \( \sqrt{3} \).

### Step 2: Calculate the hypotenuse
Using the Pythagorean theorem:
\[ \text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2 \]
Substituting the known values:
\[ \text{hypotenuse}^2 = 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \]
Thus, the hypotenuse is:
\[ \text{hypotenuse} = \sqrt{4} = 2 \]

### Step 3: Find \(\sin \theta\) and \(\cos \theta\)
Now we can find \(\sin \theta\) and \(\cos \theta\):
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{2} \]
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{3}}{2} \]

### Step 4: Calculate \(\sin^2 \theta\) and \(\cos^2 \theta\)
Now we square both values:
\[ \sin^2 \theta = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ \cos^2 \theta = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

### Step 5: Substitute into the equation
Now substitute \(\sin^2 \theta\) and \(\cos^2 \theta\) into the equation:
\[ 7 \sin^2 \theta + 3 \cos^2 \theta = 7 \left(\frac{1}{4}\right) + 3 \left(\frac{3}{4}\right) \]
Calculating each term:
\[ = \frac{7}{4} + \frac{9}{4} \]
Combine the fractions:
\[ = \frac{7 + 9}{4} = \frac{16}{4} = 4 \]

### Conclusion
Thus, we have proved that:
\[ 7 \sin^2 \theta + 3 \cos^2 \theta = 4 \]