If `tan theta= (2)/(3)` then what is the value of `(15 sin^(2) theta- 3 cos^(2)theta)/(5 sin^(2) theta+ 3 cos^(2)theta)`?

Options

A. `(33)//(32)`
B. `(11)//(29)`
C. `(33)//(47)`
D. `(33)//(45)`

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{15 \sin^2 \theta - 3 \cos^2 \theta}{5 \sin^2 \theta + 3 \cos^2 \theta} \]

given that \(\tan \theta = \frac{2}{3}\).

### Step 1: Find \(\sin^2 \theta\) and \(\cos^2 \theta\)

Since \(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{2}{3}\), we can represent \(\sin \theta\) and \(\cos \theta\) in terms of a right triangle. Let:

- Opposite side = 2
- Adjacent side = 3

Using the Pythagorean theorem, we find the hypotenuse:

\[ \text{Hypotenuse} = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

Now we can find \(\sin \theta\) and \(\cos \theta\):

\[ \sin \theta = \frac{2}{\sqrt{13}}, \quad \cos \theta = \frac{3}{\sqrt{13}} \]

Now, squaring both:

\[ \sin^2 \theta = \left(\frac{2}{\sqrt{13}}\right)^2 = \frac{4}{13}, \quad \cos^2 \theta = \left(\frac{3}{\sqrt{13}}\right)^2 = \frac{9}{13} \]

### Step 2: Substitute \(\sin^2 \theta\) and \(\cos^2 \theta\) into the expression

Now substitute \(\sin^2 \theta\) and \(\cos^2 \theta\) into the expression:

\[ \frac{15 \left(\frac{4}{13}\right) - 3 \left(\frac{9}{13}\right)}{5 \left(\frac{4}{13}\right) + 3 \left(\frac{9}{13}\right)} \]

### Step 3: Simplify the numerator and denominator

**Numerator:**

\[ 15 \cdot \frac{4}{13} - 3 \cdot \frac{9}{13} = \frac{60}{13} - \frac{27}{13} = \frac{60 - 27}{13} = \frac{33}{13} \]

**Denominator:**

\[ 5 \cdot \frac{4}{13} + 3 \cdot \frac{9}{13} = \frac{20}{13} + \frac{27}{13} = \frac{20 + 27}{13} = \frac{47}{13} \]

### Step 4: Combine the results

Now we can combine the results from the numerator and denominator:

\[ \frac{\frac{33}{13}}{\frac{47}{13}} = \frac{33}{47} \]

### Final Answer

Thus, the value of the expression is:

\[ \frac{33}{47} \]