Question

15:

15. If tan A & tan B are the roots of the quadratic equation, ax² + bx + c =0 then evaluate a sin² (A + B) + b sin (A + B). cos (A + B) + c cos² (A + B).

Answer 1

$$\begin{gathered} a{x}^2+bx+c=0 \\ Since\tan Aand\tan Barerootsoftheaboveequation \\ \tan A+\tan B=-\frac{b}{a} \\ \tan A\tan B=\frac{c}{a} \\ \tan \left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B} \\ =\frac{-\frac{b}{a}}{1-\frac{c}{a}} \\ =\frac{-\frac{b}{a}}{\frac{a-c}{a}} \\ =\frac{\frac{b}{a}}{\frac{c-a}{a}} \\ =\frac{b}{c-a} \\ a{\sin }^2\left(A+B\right)+b\sin \left(A+B\right)\cos \left(A+B\right)+c{\cos }^2\left(A+B\right) \\ ={\cos }^2\left(A+B\right)\left\{a\mathrm{ta}{n}^2\left(A+B\right)+b\tan \left(A+B\right)+c\right\} \\ =\frac{a{\tan }^2\left(A+B\right)+b\tan \left(A+B\right)+c}{se{c}^2\left(A+B\right)} \\ Usese{c}^{2}x=1+{\tan }^{2}x \\ =\frac{a{\tan }^2\left(A+B\right)+b\tan \left(A+B\right)+c}{1+{\tan }^2\left(A+B\right)} \\ =\frac{a{\left(\frac{b}{c-a}\right)}^2+b\left(\frac{b}{c-a}\right)+c}{1+{\left(\frac{b}{c-a}\right)}^2} \\ =\frac{\frac{a{b}^2}{{\left(c-a\right)}^2}+\frac{b^2}{c-a}+c}{1+\frac{b^2}{{\left(c-a\right)}^2}} \\ =\frac{\frac{a{b}^2}{{\left(c-a\right)}^2}+\frac{b^2\left(c-a\right)}{{\left(c-a\right)}^2}+{\displaystyle \frac{c\left(c-a\right){}^2}{\left(c-a\right){}^2}}}{\frac{{\left(c-a\right)}^2+b^2}{{\left(c-a\right)}^2}} \\ =\frac{a{b}^2+b^2\left(c-a\right)+{\displaystyle c\left(c-a\right){}^2}}{{\left(c-a\right)}^2+b^2} \\ =\frac{a{b}^2+b^{2}c-a{b}^2+{\displaystyle c\left(c^2+a^2-2ac\right)}}{c^2+a^2-2ac+b^2} \\ =\frac{b^{2}c+{\displaystyle c\left(c^2+a^2-2ac\right)}}{c^2+a^2-2ac+b^2} \\ =c\left(\frac{b^2+{\displaystyle c^2+a^2-2ac}}{c^2+a^2-2ac+b^2}\right) \\ =c\left(\frac{{\displaystyle c^2+a^2-2ac+b^2}}{c^2+a^2-2ac+b^2}\right) \\ =c \end{gathered}$$