If $7\cos^2 heta + 3\sin^2 heta = 4 ext{ and } 0

Q: If $7\cos^2\theta + 3\sin^2\theta = 4 \text{ and } 0 < \theta < \frac{\pi}{2}$, then what is the value of tan $\theta$ ?

To solve the problem and confirm why Option 4 is the correct answer, we need to find the value of $\tan \theta$ given the equation: \[ 7\cos^2\theta + 3\sin^2\theta = 4 \] We know the trigonometric identity: \[ \cos^2\theta + \sin^2\theta = 1 \] Let's express $\cos^2\theta$ in terms of $\sin^2\theta$: \[ \cos^2\theta = 1 - \sin^2\theta \] Substitute this into the given equation: \[ 7(1 - \sin^2\theta) + 3\sin^2\theta = 4 \] Simplify the equation: \[ 7 - 7\sin^2\theta + 3\sin^2\theta = 4 \] Combine like terms: \[ 7 - 4\sin^2\theta = 4 \] Rearrange the equation to solve for $\sin^2\theta$: \[ 7 - 4 = 4\sin^2\theta \] \[ 3 = 4\sin^2\theta \] \[ \sin^2\theta = \frac{3}{4} \] Now, find $\cos^2\theta$: \[ \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{3}{4} = \frac{1}{4} \] Since $0 < \theta < \frac{\pi}{2}$, both $\sin\theta$ and $\cos\theta$ are positive. Therefore: \[ \sin\theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \] \[ \cos\theta = \sqrt{\frac{1}{4}} = \frac{1}{2} \] Now, calculate $\tan\theta$: \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \] Thus, the value of $\tan\theta$ is $\sqrt{3}$, which corresponds to Option 4. **Explanation for Other Options:** 1. **Option 1: $\sqrt{7}$** - This option is incorrect because the calculated value of $\tan\theta$ is $\sqrt{3}$, not $\sqrt{7}$. 2. **Option 2: $\frac{7}{3}$** - This option is incorrect because the value of $\tan\theta$ derived from the given equation is $\sqrt{3}$, not $\frac{7}{3}$. 3. **Option 3: 3** - This option is incorrect because the correct value of $\tan\theta$ is $\sqrt{3}$, not 3. Therefore, the correct answer is indeed Option 4: $\sqrt{3}$.

Question

If $7\cos^2	heta + 3\sin^2	heta = 4 	ext{ and } 0 < 	heta < \frac{\pi}{2}$, then what is the value of tan $	heta$ ?

Accepted Answer

To solve the problem and confirm why Option 4 is the correct answer, we need to find the value of $\tan \theta$ given the equation:

$$ 7\cos^2\theta + 3\sin^2\theta = 4 $$

We know the trigonometric identity:

$$ \cos^2\theta + \sin^2\theta = 1 $$

Let's express $\cos^2\theta$ in terms of $\sin^2\theta$:

$$ \cos^2\theta = 1 - \sin^2\theta $$

Substitute this into the given equation:

$$ 7(1 - \sin^2\theta) + 3\sin^2\theta = 4 $$

Simplify the equation:

$$ 7 - 7\sin^2\theta + 3\sin^2\theta = 4 $$

Combine like terms:

$$ 7 - 4\sin^2\theta = 4 $$

Rearrange the equation to solve for $\sin^2\theta$:

$$ 7 - 4 = 4\sin^2\theta $$

$$ 3 = 4\sin^2\theta $$

$$ \sin^2\theta = \frac{3}{4} $$

Now, find $\cos^2\theta$:

$$ \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{3}{4} = \frac{1}{4} $$

Since $0 < \theta < \frac{\pi}{2}$, both $\sin\theta$ and $\cos\theta$ are positive. Therefore:

$$ \sin\theta = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} $$

$$ \cos\theta = \sqrt{\frac{1}{4}} = \frac{1}{2} $$

Now, calculate $\tan\theta$:

$$ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} $$

Thus, the value of $\tan\theta$ is $\sqrt{3}$, which corresponds to Option 4.

**Explanation for Other Options:**

1. **Option 1: $\sqrt{7}$** - This option is incorrect because the calculated value of $\tan\theta$ is $\sqrt{3}$, not $\sqrt{7}$.

2. **Option 2: $\frac{7}{3}$** - This option is incorrect because the value of $\tan\theta$ derived from the given equation is $\sqrt{3}$, not $\frac{7}{3}$.

3. **Option 3: 3** - This option is incorrect because the correct value of $\tan\theta$ is $\sqrt{3}$, not 3.

Therefore, the correct answer is indeed Option 4: $\sqrt{3}$.