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

We begin with the given trigonometric equation

$$ (k+1)\tan^2x-\sqrt{2}\,\lambda \tan x = (1-k), \qquad k\neq -1, \; \lambda\in\mathbb R. $$

Let us put $$t=\tan x.$$ Then the equation becomes a quadratic in $$t$$:

$$ (k+1)t^{2}-\sqrt2\,\lambda\,t-(1-k)=0. $$

The two real roots of this quadratic correspond to

$$ t_{1}= \tan\alpha,\qquad t_{2}= \tan\beta. $$

For a quadratic $$at^{2}+bt+c=0,$$ the standard Vieta relations state

$$\text{Sum of roots}= -\dfrac{b}{a},\qquad\text{Product of roots}= \dfrac{c}{a}.$$

Here we have $$a=k+1,\; b=-\sqrt2\,\lambda,\; c=-(1-k),$$ so

$$ S=t_{1}+t_{2}= -\frac{-\sqrt2\,\lambda}{\,k+1\,}= \frac{\sqrt2\,\lambda}{k+1}, $$

$$ P=t_{1}t_{2}= \frac{-(1-k)}{\,k+1\,}= \frac{k-1}{k+1}. $$

We are given the extra information

$$\tan^{2}(\alpha+\beta)=50.$$

Using the tangent-addition formula,

$$ \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\,\tan\beta} =\frac{S}{1-P}. $$

Therefore

$$ \tan^{2}(\alpha+\beta)=\left(\frac{S}{1-P}\right)^{2}=50. $$

Substituting the expressions for $$S$$ and $$P$$ we get

$$ \left(\frac{\dfrac{\sqrt2\,\lambda}{k+1}}{\,1-\dfrac{k-1}{k+1}\,}\right)^{2}=50. $$

First calculate the denominator:

$$ 1-P = 1-\frac{k-1}{k+1} = \frac{k+1}{k+1}-\frac{k-1}{k+1} = \frac{(k+1)-(k-1)}{k+1} = \frac{2}{k+1}. $$

Hence

$$ \tan^{2}(\alpha+\beta)=\left(\frac{\sqrt2\,\lambda}{k+1}\,\frac{k+1}{2}\right)^{2} =\left(\frac{\sqrt2\,\lambda}{2}\right)^{2}=50. $$

Simplifying the square gives

$$ \left(\frac{\sqrt2\,\lambda}{2}\right)^{2} =\frac{2\lambda^{2}}{4} =\frac{\lambda^{2}}{2} =50. $$

Multiplying by 2,

$$ \lambda^{2}=100. $$

So

$$ \lambda=\pm 10. $$

The question asks for “a value of $$\lambda$$,” and among the options the only value that appears is $$10.$$ Therefore we select that.

Hence, the correct answer is Option B.