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Step-by-Step Solution
Step 1: Identify the quadratic equation in tan x
The given equation is
$$(k + 1)\tan^2 x \;-\;\sqrt{2}\,\lambda \tan x \;=\;(1 - k), \quad k \neq -1.$$
Let $t = \tan x$. Then the quadratic equation becomes:
$$(k + 1)\,t^2 \;-\;\sqrt{2}\,\lambda\,t \;-\;(1 - k) = 0.$$
Step 2: Express sum and product of roots in terms of parameters
If $t = \tan \alpha$ and $t = \tan \beta$ are the two real roots, then from the quadratic equation we have:
(i) Sum of roots:
$$\tan \alpha + \tan \beta \;=\;\frac{\sqrt{2}\,\lambda}{k + 1}.$$
(ii) Product of roots:
$$\tan \alpha \,\tan \beta \;=\;\frac{-(1 - k)}{k + 1} \;=\;\frac{k - 1}{k + 1}.$$
Step 3: Formula for $\tan(\alpha + \beta)$
Recall the identity:
$$\tan(\alpha + \beta) \;=\;\frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}.$$
Substitute the sum and product of roots:
$$\tan(\alpha + \beta)
\;=\;
\frac{\dfrac{\sqrt{2}\,\lambda}{k + 1}}
{1 - \dfrac{k - 1}{k + 1}}
\;=\;
\frac{\dfrac{\sqrt{2}\,\lambda}{k + 1}}
{\dfrac{k + 1 - (k - 1)}{k + 1}}
\;=\;
\frac{\dfrac{\sqrt{2}\,\lambda}{k + 1}}
{\dfrac{2}{k + 1}}
\;=\;
\frac{\sqrt{2}\,\lambda}{2}.
$$
Simplifying, we get
$$\tan(\alpha + \beta) = \frac{\lambda}{\sqrt{2}}.$$
Step 4: Use the given condition to find $\lambda$
We are given
$$\tan^2(\alpha + \beta) = 50.$$
From the previous step,
$$\tan(\alpha + \beta) = \frac{\lambda}{\sqrt{2}} \quad \Longrightarrow \quad
\left(\frac{\lambda}{\sqrt{2}}\right)^2 = 50.$$
Hence,
$$\frac{\lambda^2}{2} = 50 \quad \Longrightarrow \quad \lambda^2 = 100.$$
Since $\lambda$ is a real number,
$$\lambda = 10 \quad \text{or} \quad \lambda = -10.$$
A possible value of $\lambda$ consistent with the given options is
$$\boxed{10}.$$
