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(B) The given equation is $(k+1) 	an^2 x - (\sqrt{2} \lambda) 	an x + (k-1) = 0$.Let $t = 	an x$. The roots of the quadratic equation $(k+1)t^2 - (

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$10$

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(B) The given equation is $(k+1) 	an^2 x - (\sqrt{2} \lambda) 	an x + (k-1) = 0$.Let $t = 	an x$. The roots of the quadratic equation $(k+1)t^2 - (\sqrt{2} \lambda)t + (k-1) = 0$ are $	an \alpha$ and $	an \beta$.From the properties of quadratic equations,the sum of roots is $	an \alpha + 	an \beta = \frac{\sqrt{2} \lambda}{k+1}$ and the product of roots is $	an \alpha 	an \beta = \frac{k-1}{k+1}$.Using the tangent addition formula,$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Substituting the values,$	an(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k+1}}{1 - \frac{k-1}{k+1}} = \frac{\sqrt{2} \lambda}{k+1 - k + 1} = \frac{\sqrt{2} \lambda}{2} = \frac{\lambda}{\sqrt{2}}$.Given $	an^2(\alpha + \beta) = 50$,we have $\left(\frac{\lambda}{\sqrt{2}}ight)^2 = 50$.$\frac{\lambda^2}{2} = 50 \Rightarrow \lambda^2 = 100 \Rightarrow \lambda = \pm 10$.Thus,a possible value of $\lambda$ is $10$.

Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1) \tan ^{2} x-\sqrt{2} \lambda \tan x=(1-k)$ where $k(\neq-1)$ and $\lambda$ are real numbers. If $\tan ^{2}(\alpha+\beta)=50,$ then a value of $\lambda$ is :

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