If `2tan""(alpha)/(2) = tan ""(beta)/(2)`, then `(3+5cosbeta)/(5+3cosbeta)` is equal to :

Options

A. `cos alpha`
B. `cos beta`
C. `sin alpha`
D. `sin beta`

Answer

To solve the problem, we start with the given equation:

\[ \frac{2 \tan(\frac{\alpha}{2})}{2} = \frac{\tan(\frac{\beta}{2})}{2} \]

This simplifies to:

\[ \tan(\frac{\alpha}{2}) = \frac{1}{2} \tan(\frac{\beta}{2}) \]

Next, we need to find the value of:

\[ \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta} \]

Using the identity for cosine in terms of tangent, we can express \(\cos \beta\) in terms of \(\tan(\frac{\beta}{2})\):

\[ \cos \beta = \frac{1 - \tan^2(\frac{\beta}{2})}{1 + \tan^2(\frac{\beta}{2})} \]

Let \(T = \tan^2(\frac{\beta}{2})\). Then:

\[ \cos \beta = \frac{1 - T}{1 + T} \]

Substituting this into our expression gives:

\[ \frac{3 + 5 \left(\frac{1 - T}{1 + T}\right)}{5 + 3 \left(\frac{1 - T}{1 + T}\right)} \]

This simplifies to:

\[ \frac{3(1 + T) + 5(1 - T)}{5(1 + T) + 3(1 - T)} \]

Expanding both the numerator and the denominator:

Numerator:

\[ 3 + 3T + 5 - 5T = 8 - 2T \]

Denominator:

\[ 5 + 5T + 3 - 3T = 8 + 2T \]

Thus, we have:

\[ \frac{8 - 2T}{8 + 2T} \]

Since we know from the problem statement that:

\[ 2 \tan(\frac{\alpha}{2}) = \tan(\frac{\beta}{2}) \]

Squaring both sides gives:

\[ 4 \tan^2(\frac{\alpha}{2}) = \tan^2(\frac{\beta}{2}) \implies T = 4 \tan^2(\frac{\alpha}{2}) \]

Substituting \(T\) back into our expression:

\[ \frac{8 - 2(4 \tan^2(\frac{\alpha}{2}) )}{8 + 2(4 \tan^2(\frac{\alpha}{2}) )} \]

This simplifies to:

\[ \frac{8 - 8 \tan^2(\frac{\alpha}{2})}{8 + 8 \tan^2(\frac{\alpha}{2})} = \frac{8(1 - \tan^2(\frac{\alpha}{2}))}{8(1 + \tan^2(\frac{\alpha}{2}))} \]

Cancelling out the 8 gives:

\[ \frac{1 - \tan^2(\frac{\alpha}{2})}{1 + \tan^2(\frac{\alpha}{2})} \]

Using the identity for cosine:

\[ \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)} = \cos(2\theta) \]

Thus, we have:

\[ \cos(\alpha) \]

Therefore, the final answer is:

\[ \frac{3 + 5 \cos \beta}{5 + 3 \cos \beta} = \cos(\alpha) \]