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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 :
Option: 1 $5\sqrt{2}$
Option: 2 $10\sqrt{2}$
Option: 3 $10$
Option: 4 $5$

Answers (1)

best_answer

As we have learnt,

Sum of roots:

$\\\mathrm{\alpha + \beta =\frac{-b}{a}}$

Product of roots:

$\alpha \cdot \beta = \frac{c}{a}$

Trigonometric Ratio for Compound Angles (Part 2)

$\\\mathrm{\tan (\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}}\\\\\mathrm{\tan (\alpha-\beta)=\frac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}}$

Now,

$\\\tan \alpha + \tan{\beta } = \frac{\sqrt{2} \lambda}{1+k}\\\\\tan \alpha \times \tan{\beta } = \frac{k-1}{1+k}$

Since $\tan \alpha \& \tan \beta$ are the roots of the given equation

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan{\beta }}{1-\tan{\alpha }\tan{\beta }}$ $= \frac{\frac{\sqrt{2}\lambda }{1+k}}{1- \frac{k-1}{k+1}} =\frac{\lambda}{\sqrt2}$

Now,

$\\ {\tan ^{2}(\alpha+\beta)=\frac{\lambda^{2}}{2}=50} \\ {\lambda=10}$

Posted by

Kuldeep Maurya

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