You have provided the following:

1. $3 \tan(a) = 4$, which implies $\tan(a) = \frac{4}{3}$.

We are tasked to find the value of:

$\frac{3 \sin(a) + 4 \cos(a)}{6 \sin(a) - 5 \cos(a)}.$

Step-by-step solution:

Step 1: Find $\sin(a)$ and $\cos(a)$ using $\tan(a)$.

Since $\tan(a) = \frac{4}{3}$, we know that:

$\tan(a) = \frac{\sin(a)}{\cos(a)}.$

From $\tan(a) = \frac{4}{3}$, we can assign:

• $\sin(a) = 4k$
• $\cos(a) = 3k$,

for some constant $k$.

Step 2: Use the identity $\sin^2(a) + \cos^2(a) = 1$.

Substitute $\sin(a) = 4k$ and $\cos(a) = 3k$ into this identity:

$(4k)^2 + (3k)^2 = 1,$ $16k^2 + 9k^2 = 1,$ $25k^2 = 1,$ $k^2 = \frac{1}{25},$ $k = \frac{1}{5}.$

Thus, we have:

• $\sin(a) = \frac{4}{5}$
• $\cos(a) = \frac{3}{5}$.

Step 3: Substitute the values of $\sin(a)$ and $\cos(a)$ into the expression.

Now, substitute $\sin(a) = \frac{4}{5}$ and $\cos(a) = \frac{3}{5}$ into:

$\frac{3 \sin(a) + 4 \cos(a)}{6 \sin(a) - 5 \cos(a)}.$

The numerator becomes:

$3 \sin(a) + 4 \cos(a) = 3 \times \frac{4}{5} + 4 \times \frac{3}{5} = \frac{12}{5} + \frac{12}{5} = \frac{24}{5}.$

The denominator becomes:

$6 \sin(a) - 5 \cos(a) = 6 \times \frac{4}{5} - 5 \times \frac{3}{5} = \frac{24}{5} - \frac{15}{5} = \frac{9}{5}.$

Thus, the expression simplifies to:

$\frac{\frac{24}{5}}{\frac{9}{5}} = \frac{24}{9} = \frac{8}{3}.$

Final Answer:

$\frac{3 \sin(a) + 4 \cos(a)}{6 \sin(a) - 5 \cos(a)} = \frac{8}{3}.$

Would you like more details or further explanations?

Here are 5 related questions to expand on this:

1. How do you derive $\sin(a)$ and $\cos(a)$ from $\tan(a)$?
2. What are some common trigonometric identities and how are they used in problems like this?
3. Can you solve similar expressions for different values of $\tan(a)$?
4. How does the Pythagorean identity $\sin^2(a) + \cos^2(a) = 1$ help in simplifying expressions?
5. Can you explain how the unit circle relates to $\sin(a)$ and $\cos(a)$?

Tip: When solving trigonometric equations, always look for ways to simplify using identities like $\sin^2(a) + \cos^2(a) = 1$. This helps you relate $\sin(a)$ and $\cos(a)$ effectively.