Let $\displaystyle -\frac { \pi }{ 6 } <\theta <-\frac { \pi }{ 12 } $, Suppose $\displaystyle { \alpha }{ 1 }$ and $\displaystyle { \beta }{ 1 }$ are the roots of the equation $\displaystyle { x }^{ 2 }-2x\sec { \theta } +1=0$ and $\displaystyle { \alpha }{ 2 }$ and $\displaystyle { \beta }{ 2 }$ are the roots of the equation $\displaystyle { x }^{ 2 }+2x\tan { \theta } -1=0$. If $\displaystyle { \alpha }{ 1 }>{ \beta }{ 1 }$ and $\displaystyle { \alpha }{ 2 }>{ \beta }{ 2 }$, then $\displaystyle { \alpha }{ 1 }+{ \beta }{ 2 }$ equals to

For the first equation x^2 - 2xsec(theta) + 1 = 0, the product of the roots is 1, meaning the roots are reciprocals of each other. Since alpha_1 is greater than beta_1, beta_1 is the smaller positive root, which evaluates to sec(theta) minus the square root of sec^2(theta) minus 1. Using the trigonometric identity sec^2(theta) minus 1 equals tan^2(theta), this root simplifies to sec(theta) minus tan(theta). For the second equation, the sum of the roots alpha_2 and beta_2 equals negative 2tan(theta) by Vieta's formulas, and since the roots are of opposite signs and alpha_2 is greater than beta_2, alpha_2 is the positive root, making beta_2 the negative root equal to negative tan(theta) plus sec(theta). Adding alpha_1 and beta_2 gives the result negative 2tan(theta).

• Thanks
• Wrong explanation