If `alpha,beta in (-pi/2,0)` such that `(sin alpha+sinbeta)+(sinalpha)/(sinbeta)=0 and (sinalpha+sinbeta) (sinalpha)/(sinbeta)=-1 and lambda=lim_(n->oo) (1+(2sinalpha)^(2n))/((2sinbeta)^(2n))` then :

Options

A. `alpha = (pi)/(6)`
B. `lamda =2`
C. `alpha =- (pi)/(3)`
D. `lamda =1`

Answer

To solve the problem step by step, we will analyze the given equations and find the limit as specified.

### Step 1: Analyze the given equations

We have two equations:

1. \((\sin \alpha + \sin \beta) + \frac{\sin \alpha}{\sin \beta} = 0\)
2. \((\sin \alpha + \sin \beta) \cdot \frac{\sin \alpha}{\sin \beta} = -1\)

### Step 2: Rearranging the first equation

From the first equation, we can rearrange it as follows:

\[ \sin \alpha + \sin \beta = -\frac{\sin \alpha}{\sin \beta} \]

### Step 3: Substitute into the second equation

Now, substitute \(\sin \alpha + \sin \beta\) from the first equation into the second equation:

\[ \left(-\frac{\sin \alpha}{\sin \beta}\right) \cdot \frac{\sin \alpha}{\sin \beta} = -1 \]

This simplifies to:

\[ -\frac{\sin^2 \alpha}{\sin^2 \beta} = -1 \]

### Step 4: Simplifying the equation

Removing the negative sign gives us:

\[ \frac{\sin^2 \alpha}{\sin^2 \beta} = 1 \]

This implies:

\[ \sin^2 \alpha = \sin^2 \beta \]

### Step 5: Finding the relationship between \(\alpha\) and \(\beta\)

Since \(\alpha\) and \(\beta\) are in the interval \((- \frac{\pi}{2}, 0)\), we conclude:

\[ \sin \alpha = \sin \beta \implies \alpha = \beta \]

### Step 6: Choosing a specific value for \(\alpha\) and \(\beta\)

Let’s choose \(\alpha = \beta = -\frac{\pi}{6}\) (as suggested in the video).

### Step 7: Calculate \(\lambda\)

Now we need to find:

\[ \lambda = \lim_{n \to \infty} \frac{1 + (2 \sin \alpha)^{2n}}{(2 \sin \beta)^{2n}} \]

Substituting \(\alpha = \beta = -\frac{\pi}{6}\):

\[ \sin \left(-\frac{\pi}{6}\right) = -\frac{1}{2} \]

Thus,

\[ 2 \sin \alpha = 2 \left(-\frac{1}{2}\right) = -1 \]
\[ 2 \sin \beta = 2 \left(-\frac{1}{2}\right) = -1 \]

Now substituting these values into \(\lambda\):

\[ \lambda = \lim_{n \to \infty} \frac{1 + (-1)^{2n}}{(-1)^{2n}} = \lim_{n \to \infty} \frac{1 + 1}{1} = 2 \]

### Final Answer

Thus, the value of \(\lambda\) is:

\[ \lambda = 2 \]