Let -pi/6<theta<-pi/12. Suppose a1 and b1 are the roots of the equation x² - 2xsec(theta) + 1=0 and a2 and b2 are the roots of the equation x² + 2xtan(theta) - 1=0. If a1>b1 and a2>b2, then a1 + b2 is???

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Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{x}}^2-2\mathrm{x}\left(\mathrm{sec\theta }\right)+1=0 \\ \mathrm{Using} \mathrm{Shridharacharya}\text{'}\mathrm{s} \mathrm{formula} \mathrm{we} \mathrm{get}: \\ \mathrm{x}=\frac{-\left(-2\mathrm{sec\theta }\right)\pm \sqrt{{\left(-2\mathrm{sec\theta }\right)}^2-4\times 1\times 1}}{2\times 1} \\ =\frac{2\mathrm{sec\theta }\pm \sqrt{4{\sec }^2\mathrm{\theta }-4}}{2} \\ =\frac{2\mathrm{sec\theta }\pm 2\sqrt{{\sec }^2\mathrm{\theta }-1}}{2} \\ =\mathrm{sec\theta }\pm \sqrt{{\tan }^2\mathrm{\theta }} \\ =\mathrm{sec\theta }\pm \left|\mathrm{tan\theta }\right| \\ \mathrm{As} -\frac{\mathrm{\pi }}{6}<\mathrm{\theta }<-\frac{\mathrm{\pi }}{12}, \mathrm{hence} \mathrm{tan\theta } \mathrm{will} \mathrm{be} \mathrm{negative}. \\ \Rightarrow \left|\mathrm{tan\theta }\right|=-\mathrm{tan\theta } \\ \Rightarrow \mathrm{x}=\mathrm{sec\theta }\mp \mathrm{tan\theta } \\ {\mathrm{a}}_1=\mathrm{sec\theta }-\mathrm{tan\theta } \\ {\mathrm{a}}_2=\mathrm{sec\theta }+\mathrm{tan\theta } \\ {\mathrm{x}}^2+2\mathrm{x}\left(\mathrm{tan\theta }\right)-1=0 \\ \mathrm{x}=\frac{-\left(2\mathrm{tan\theta }\right)\pm \sqrt{{\left(-2\mathrm{tan\theta }\right)}^2-4\times 1\times 1\left(-1\right)}}{2\times 1} \\ =\frac{-2\mathrm{tan\theta }\pm \sqrt{4{\tan }^2\mathrm{\theta }+4}}{2} \\ =\frac{-2\mathrm{tan\theta }\pm 2\sqrt{{\tan }^2\mathrm{\theta }+1}}{2} \\ =\frac{-2\mathrm{tan\theta }\pm 2\sqrt{{\sec }^2\mathrm{\theta }}}{2} \\ =-\mathrm{tan\theta }\pm \left|\mathrm{sec\theta }\right| \\ \mathrm{As} \mathrm{sec\theta } \mathrm{is} \mathrm{positive} \mathrm{when} -\frac{\mathrm{\pi }}{6}<\mathrm{\theta }<-\frac{\mathrm{\pi }}{12} \\ \Rightarrow \left|\mathrm{sec\theta }\right|=\mathrm{sec\theta } \\ \Rightarrow \mathrm{x}=-\mathrm{tan\theta }\pm \mathrm{sec\theta }, \mathrm{hence} \mathrm{roots} \mathrm{are} \\ \mathrm{x}=\mathrm{sec\theta }-\mathrm{tan\theta } \mathrm{and} \mathrm{x}=-\mathrm{sec\theta }-\mathrm{tan\theta } \\ \mathrm{As} \\ {\mathrm{a}}_1=\mathrm{sec\theta }-\mathrm{tan\theta } \\ {\mathrm{a}}_2=\mathrm{sec\theta }+\mathrm{tan\theta } \\ \mathrm{Note} \mathrm{that} \mathrm{sec\theta }-\mathrm{tan\theta } \mathrm{is} \mathrm{positive} \\ \mathrm{and} {\mathrm{a}}_1>{\mathrm{b}}_1 \mathrm{and} {\mathrm{a}}_2>{\mathrm{b}}_2 \\ \Rightarrow {\mathrm{b}}_1=-\mathrm{sec\theta }-\mathrm{tan\theta } \\ \mathrm{and} \\ {\mathrm{b}}_2=\mathrm{sec\theta }-\mathrm{tan\theta } \\ \mathrm{But} \mathrm{in} \mathrm{that} \mathrm{case} \\ {\mathrm{b}}_2>{\mathrm{a}}_2 \\ \mathrm{as} \\ \mathrm{sec\theta }-\mathrm{tan\theta }>-\mathrm{sec\theta }-\mathrm{tan\theta } \\ \mathrm{No} \mathrm{combination} \mathrm{fits} {\mathrm{a}}_1>{\mathrm{b}}_1 \mathrm{and} {\mathrm{a}}_2>{\mathrm{b}}_2 \\ \mathrm{You} \mathrm{can} \mathrm{also} \mathrm{see} \mathrm{that} \mathrm{in} \mathrm{the} \mathrm{graph} \mathrm{given} \mathrm{below}. \end{gathered}$$


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