Please solve:

153. If sec $\alpha$ and cosec $\alpha$ are the roots of the equation x² -ax +b =0, then

(a) a² = b(b -2)                   (b) a² =b (b +2)  ​
(c) a² + b² =2b                    (d) none of these

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{x}}^2-\mathrm{ax}+\mathrm{b}=0 \\ \mathrm{Roots} \mathrm{are} \mathrm{sec\alpha } \mathrm{and} \mathrm{cosec\alpha } \\ \mathrm{Product} \mathrm{of} \mathrm{roots}=\mathrm{b} \\ \Rightarrow \mathrm{sec\alpha }.\mathrm{cosec\alpha }=\mathrm{b} \\ \Rightarrow \frac{1}{\mathrm{cos\alpha }}\frac{1}{\mathrm{sin\alpha }}=\mathrm{b} \\ \Rightarrow \mathrm{sin\alpha }.\mathrm{cos\alpha }=\frac{1}{\mathrm{b}}...\left(\mathrm{i}\right) \\ \mathrm{Sum} \mathrm{of} \mathrm{roots}=\mathrm{a} \\ \Rightarrow \mathrm{sec\alpha } +\mathrm{cosec\alpha }=\mathrm{a} \\ \Rightarrow \frac{1}{\mathrm{cos\alpha }}+\frac{1}{\mathrm{sin\alpha }}=\mathrm{a} \\ \Rightarrow \frac{\mathrm{sin\alpha }+\mathrm{cos\alpha }}{\mathrm{sin\alpha }.\mathrm{cos\alpha }}=\mathrm{a} \\ \Rightarrow \mathrm{sin\alpha }+\mathrm{cos\alpha }=\mathrm{asin\alpha }.\mathrm{cos\alpha } \\ \Rightarrow \mathrm{sin\alpha }+\mathrm{cos\alpha }=\frac{\mathrm{a}}{\mathrm{b}} \left[\mathrm{Use} \left(\mathrm{i}\right)\right] \\ \mathrm{Square} \mathrm{both} \mathrm{sides} \\ \Rightarrow {\left(\mathrm{sin\alpha }+\mathrm{cos\alpha }\right)}^2=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2} \\ \Rightarrow {\sin }^2\mathrm{\alpha }+{\cos }^2\mathrm{\alpha }+2\mathrm{sin\alpha }.\mathrm{cos\alpha }=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2} \\ \Rightarrow 1+\frac{2}{\mathrm{b}}=\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2} \\ \Rightarrow {\mathrm{b}}^2+{\mathrm{b}}^2\frac{2}{\mathrm{b}}={\mathrm{a}}^2 \\ \Rightarrow {\mathrm{a}}^2={\mathrm{b}}^2+2\mathrm{b} \\ \Rightarrow {\mathrm{a}}^2=\mathrm{b}\left(\mathrm{b}+2\right)...\mathrm{Option}\left(\mathrm{b}\right) \end{gathered}$$

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