Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1) \tan ^{2} x-\sqrt{2} \cdot \lambda \tan x=(1-k)$, where $k(\neq-1)$ and $\lambda$ are real numbers. If $\tan ^{2}(\alpha+\beta)=50$, then a value of $\lambda$ is:
Correct Option: , 2
$(k+1) \tan ^{2} x-\sqrt{2} \lambda \tan x+(k-1)=0$
$\tan \alpha+\tan \beta=\frac{\sqrt{2} \lambda}{k+1}$ [Sum of roots]
$\tan \alpha \cdot \tan \beta=\frac{k-1}{k+1}$ [Product of roots]
$\therefore \quad \tan (\alpha+\beta)=\frac{\frac{\sqrt{2} \lambda}{k+1}}{1-\frac{k-1}{k+1}}=\frac{\sqrt{2} \lambda}{2}=\frac{\lambda}{\sqrt{2}}$
$\tan ^{2}(\alpha+\beta)=\frac{\lambda^{2}}{2}=50$
$\lambda=10$
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