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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions
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In a triangle ABC with $\angle C=90^{\large\circ}$ the equation whose roots are $\tan A$ and $\tan B$ is ___________

$\begin{array}{1 1}(A)\;x^2-\large\frac{2}{\sin 2A}\normalsize x+1\\(B)\;x^2+\large\frac{2}{\sin 2A}\normalsize x+1\\(C)\;x^2-\large\frac{2}{\sin A}\normalsize x+1\\(D)\;x^2-\large\frac{1}{\sin 2A}\normalsize x+1\end{array} $

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1 Answer

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Toolbox:
  • $\tan A+\tan B=\tan A+\large\frac{1}{\tan A}=\frac{2\tan^2+1}{2\tan A}=\frac{2}{\sin 2A}$
  • $\sin 2A=\large\frac{2\tan A}{1+\tan^2A}$
$\angle C=90^{\large\circ}$
$\angle A+\angle B=90^{\large\circ}$
$\Rightarrow \tan A.\tan B =1$
$\tan B=\large\frac{1}{\tan A}$
$\tan A$ and $\tan B$ are roots
$(x-\tan A)(x-\tan B)$
$\Rightarrow x^2-(\tan A+\tan B)x+\tan A\tan B$
$\Rightarrow x^2-(\tan A+\tan B)x+1$
$\tan A+\tan B=\tan A+\large\frac{1}{\tan A}=\frac{2\tan^2+1}{2\tan A}=\frac{2}{\sin 2A}$
$\sin 2A=\large\frac{2\tan A}{1+\tan^2A}$
$\Rightarrow x^2-\large\frac{2}{\sin 2A}$$x+1$
Hence (A) is the correct answer.
answered May 20, 2014 by sreemathi.v
 

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