Let -frac{pi}{6}

Question

(C) Given $-\frac{\pi}{6} < 	heta < -\frac{\pi}{12}$.For the equation $x^2 - 2x \sec 	heta + 1 = 0$,the roots are $\alpha_1, \beta_1 = \frac{2 \sec

Vedclass · Accepted Answer

$-2 	an 	heta$

Vedclass · Answer

(C) Given $-\frac{\pi}{6} For the equation $x^2 - 2x \sec 	heta + 1 = 0$,the roots are $\alpha_1, \beta_1 = \frac{2 \sec 	heta \pm \sqrt{4 \sec^2 	heta - 4}}{2} = \sec 	heta \pm 	an 	heta$.Since $	heta \in (-\frac{\pi}{6}, -\frac{\pi}{12})$,$\sec 	heta > 0$ and $	an 	heta  \sec 	heta + 	an 	heta$. Given $\alpha_1 > \beta_1$,we have $\alpha_1 = \sec 	heta - 	an 	heta$.For the equation $x^2 + 2x 	an 	heta - 1 = 0$,the roots are $\alpha_2, \beta_2 = \frac{-2 	an 	heta \pm \sqrt{4 	an^2 	heta + 4}}{2} = -	an 	heta \pm \sec 	heta$.Since $\sec 	heta > 0$,we have $-	an 	heta + \sec 	heta > -	an 	heta - \sec 	heta$. Given $\alpha_2 > \beta_2$,we have $\beta_2 = -	an 	heta - \sec 	heta$.Therefore,$\alpha_1 + \beta_2 = (\sec 	heta - 	an 	heta) + (-	an 	heta - \sec 	heta) = -2 	an 	heta$.

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