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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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