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We have given

$\left( {k + 1} \right){\tan ^2}x - \sqrt 2 \lambda \tan x = \left( {1 - k} \right)$

We can rewrite it as :

$\left( {k + 1} \right){\tan ^2}x - \sqrt 2 \lambda \tan x + \left( {k - 1} \right) = 0$

Roots of this quadratic equation are $\tan \alpha $ and $\tan \beta $.

From the property of sum of roots and product of roots we can write

$\tan \alpha + \tan \beta = \dfrac{{\sqrt 2 \lambda }}{{\left( {k + 1} \right)}}$

$\tan \alpha \times \tan \beta = \dfrac{{\left( {k - 1} \right)}}{{\left( {k + 1} \right)}}$

And we know the formula

$\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha .\tan \beta }}$

On putting values from above we get,

$\tan \left( {\alpha + \beta } \right) = \dfrac{{\dfrac{{\sqrt 2 \lambda }}{{\left( {k + 1} \right)}}}}{{1 - \dfrac{{\left( {k - 1} \right)}}{{\left( {k + 1} \right)}}}}$

On further solving we get,

$\tan \left( {\alpha + \beta } \right) = \dfrac{{\dfrac{{\sqrt 2 \lambda }}{{\left( {k + 1} \right)}}}}{{\dfrac{{\left( {k + 1} \right) - \left( {k - 1} \right)}}{{\left( {k + 1} \right)}}}}$

On cancel out we get,

$\tan \left( {\alpha + \beta } \right) = \dfrac{{\sqrt 2 \lambda }}{2} = \dfrac{\lambda }{{\sqrt 2 }}$

And hence on squaring we get,

${\tan ^2}\left( {\alpha + \beta } \right) = \dfrac{{{\lambda ^2}}}{2}$

And we have given in the question ${\tan ^2}\left( {\alpha + \beta } \right) = 50$

On comparing both equation we get

$

\dfrac{{{\lambda ^2}}}{2} = 50 \\

{\lambda ^2} = 100 \\

\therefore \lambda = \pm 10 \\

$