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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Quadratic Equation

NDA 2016 · Mathematics · Quadratic Equation

Given that α and β are the roots of the equation x^2+b x+c=0 with b...

Given that tan⁡α\tan \alpha and tan⁡β\tan \beta are the roots of the equation x2+bx+c=0x^2+b x+c=0 with b≠0b \neq 0. What is tan⁡(α+β)\tan (\alpha+\beta) equal to?

Options

  1. Ab(c−1)\mathrm{b}(\mathrm{c}-1)
  2. Bc(b−1)c(b-1)
  3. Cc(b−1)−1c(b-1)^{-1}
  4. Db(c−1)−1b(c-1)^{-1}

Correct answer

D. b(c−1)−1b(c-1)^{-1}

Step-by-step solution

x^2+b x+c=0 \quad b \neq 0\because \tan \alphaandand\tan \betaarerootsofequation.are roots of equation.∴tan⁡α+tan⁡β=−b1=−b...(1)∴tan⁡αtan⁡β=c1=ctan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=−b1−c=b(c−1)−1\begin{aligned} & \therefore \tan \alpha+\tan \beta=\frac{-b}{1}=-b ...(1)\\ & \therefore \tan \alpha \tan \beta=\frac{c}{1}=c \\ & \begin{array}{c} \tan(\alpha+\beta)=\frac{\tan \alpha+\tan \beta}{1-\tan \alpha\tan \beta}=\frac{-b}{1-c} \\ \quad=b(c-1)^{-1} \end{array} \end{aligned}$
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