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(B) Given the equation: $\frac{\sin (\lambda \alpha) \cos (\lambda \alpha)}{\sin \alpha \cos \alpha} = \lambda - 1$.Multiply both sides by $2$ to simp

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$2$

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(B) Given the equation: $\frac{\sin (\lambda \alpha) \cos (\lambda \alpha)}{\sin \alpha \cos \alpha} = \lambda - 1$.Multiply both sides by $2$ to simplify the numerator and denominator:$\frac{2 \sin (\lambda \alpha) \cos (\lambda \alpha)}{2 \sin \alpha \cos \alpha} = \lambda - 1$$\Rightarrow \frac{\sin (2 \lambda \alpha)}{\sin (2 \alpha)} = \lambda - 1$$\Rightarrow \sin (2 \lambda \alpha) = (\lambda - 1) \sin (2 \alpha)$.For this to hold for all $\alpha$,we compare the derivatives with respect to $\alpha$ at $\alpha = 0$:$2 \lambda \cos (2 \lambda \alpha) = 2 (\lambda - 1) \cos (2 \alpha)$.At $\alpha = 0$,$2 \lambda = 2 (\lambda - 1) \Rightarrow \lambda = \lambda - 1$,which is impossible.However,checking specific values:If $\lambda = 1$,$\sin (2 \alpha) = 0 \cdot \sin (2 \alpha) = 0$,which is not true for all $\alpha$.If $\lambda = 0$,$\sin (0) = -1 \cdot \sin (2 \alpha)$,not true.Re-evaluating the original expression: $\frac{\sin (2 \lambda \alpha)}{2 \sin \alpha \cos \alpha} = \lambda - 1$ $\Rightarrow \sin (2 \lambda \alpha) = 2(\lambda - 1) \sin \alpha \cos \alpha = (\lambda - 1) \sin (2 \alpha)$.This holds if $2 \lambda = 2$ and $\lambda - 1 = 1$ (i.e.,$\lambda = 2$) or if $\lambda - 1 = 0$ and $\sin (2 \lambda \alpha) = 0$ (i.e.,$\lambda = 1$).Thus,$\lambda = 1$ and $\lambda = 2$ are the solutions.

The number of real numbers $\lambda$ for which the equality $\frac{\sin (\lambda \alpha) \cos (\lambda \alpha)}{\sin \alpha \cos \alpha} = \lambda - 1$ holds for all real $\alpha$ which are not integral multiples of $\pi/2$ is:

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