In a Triangle ABC with ∠C = π/2, Find the Equation Whose Roots Are tan A and tan B
Question:
In a triangle ABC with \[ \angle C=\frac{\pi}{2}, \] find the equation whose roots are \[ \tan A \quad \text{and} \quad \tan B. \]
Solution
Since
\[ A+B=\frac{\pi}{2} \]we have
\[ \tan(A+B)=\tan\frac{\pi}{2} \]Hence,
\[ 1-\tan A \tan B=0 \] \[ \tan A \tan B=1 \]Also,
\[ B=\frac{\pi}{2}-A \] \[ \tan B=\cot A=\frac{1}{\tan A} \]Therefore, the roots are reciprocals of each other.
Let the roots be α and β.
\[ \alpha+\beta=\tan A+\tan B \] Using \[ \tan A+\tan B = \frac{\sin(A+B)}{\cos A \cos B} = \frac{1}{\cos A \cos B} \] and \[ \tan A \tan B=1 \] we get \[ \cos A \cos B = \frac{1}{2} \] because \[ A+B=\frac{\pi}{2} \] Thus, \[ \alpha+\beta=2 \] and \[ \alpha\beta=1 \]The required quadratic equation is
\[ x^2-(\alpha+\beta)x+\alpha\beta=0 \] \[ x^2-2x+1=0 \]