In a triangle ABC with $\angle C=90^{\large\circ}$ the equation whose roots are $\tan A$ and $\tan B$ is ___________

Question

$\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} $

Answer 1

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.