Let -π/6

Q: Why do we get $\sec \theta - \tan \theta$ and not $\sec \theta + \tan \theta$ for $\alpha_1$?

Great question! We are given $\alpha_1 > \beta_1$. Since $\tan \theta$ is negative in the Fourth Quadrant, $-\tan \theta$ is positive. Thus, $\sec \theta - \tan \theta$ is larger than $\sec \theta + \tan \theta$.

Q: Can I just use the sum of roots formula $x_1 + x_2 = -b/a$?

No, the sum of roots formula gives $\alpha_1 + \beta_1 = 2 \sec \theta$. We need $\alpha_1 + \beta_2$, which involves roots from *different* equations. You must solve for them individually.

Q: Why is $\sqrt{\tan^2 \theta} = -\tan \theta$?

Because $\sqrt{x^2} = |x|$. Since $\tan \theta < 0$ in the Fourth Quadrant, $|\tan \theta| = -\tan \theta$. It is a classic trap to forget the negative sign!

Question

Let $-\pi/6 < heta < -\pi/12$. Suppose $\alpha_1$ and $\beta_1$ are the roots of the equation $x^2 - 2x \sec heta + 1 = 0$ and $\alpha_2$ and $\beta_2$ are the roots of the equation $x^2 + 2x an heta - 1 = 0$. If $\alpha_1 > \beta_1$ and $\alpha_2 > \beta_2$, then $\alpha_1 + \beta_2$ equals Options: (A) $2(\sec heta - an heta)$ (B) $2 \sec heta$ (C) $-2 an heta$ (D) $0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Welcome, fellow traveler, to the beautiful world of JEE Advanced mathematics. Today, we are not just solving a quadratic equation; we are embarking on a journey through the unit circle. Imagine you are standing on the coordinate plane, looking at the interval $-\frac{\pi}{6} < heta < -\frac{\pi}{12}$. This is our starting point, which corresponds to the region between $-30^\circ$ and $-15^\circ$. This region lies in the Fourth Quadrant ($Q4$). In the Fourth Quadrant, the cosine function is positive, which means $\sec heta > 0$, but the tangent function is negative, so $ an heta < 0$. This quadrant check is the key that unlocks the entire problem. ### The First Quadratic: The $\sec heta$ Dance Let us tackle the first equation: $x^2 - 2x \sec heta + 1 = 0$. Applying the quadratic formula, we get: $$x = \frac{2\sec heta \pm \sqrt{4\sec^2 heta - 4}}{2}$$ Simplifying this, we find $x = \sec heta \pm \sqrt{\sec^2 heta - 1}$. Using the identity $\sec^2 heta - 1 = an^2 heta$, we obtain $x = \sec heta \pm \sqrt{ an^2 heta}$. Here is where the magic happens. We know $\sqrt{ an^2 heta} = | an heta|$. Since $ an heta < 0$ in $Q4$, $| an heta| = - an heta$. Thus, our roots are $\sec heta \pm (- an heta)$. Since $\alpha_1 > \beta_1$ and $- an heta$ is positive, the larger root is: $$\alpha_1 = \sec heta - an heta$$ ### The Second Quadratic: The $ an heta$ Dance Now, let us look at the second equation: $x^2 + 2x an heta - 1 = 0$. Using the quadratic formula again, we get: $$x = \frac{-2 an heta \pm \sqrt{4 an^2 heta + 4}}{2}$$ This simplifies to $x = - an heta \pm \sqrt{ an^2 heta + 1}$. Using the identity $1 + an^2 heta = \sec^2 heta$, we get $x = - an heta \pm \sqrt{\sec^2 heta}$. Since $\sec heta > 0$ in $Q4$, $\sqrt{\sec^2 heta} = \sec heta$. The roots are $x = - an heta \pm \sec heta$. We are given $\alpha_2 > \beta_2$, so the smaller root is: $$\beta_2 = - an heta - \sec heta$$ ### The Grand Finale We have our two pieces: $\alpha_1 = \sec heta - an heta$ and $\beta_2 = - an heta - \sec heta$. Now, let us calculate the sum $\alpha_1 + \beta_2$. Substituting our expressions, we get: $$\alpha_1 + \beta_2 = (\sec heta - an heta) + (- an heta - \sec heta)$$ Look closely at the terms. The $\sec heta$ and $-\sec heta$ cancel out perfectly! This leaves us with $- an heta - an heta$, which results in: $$\alpha_1 + \beta_2 = -2 an heta$$ The complexity vanishes, leaving behind a simple, elegant result. You have successfully navigated the traps and arrived at the solution.

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Visualized Solution

Locating $θ$ on the Unit Circle

Signs of Trigonometric Functions in $Q 4$

Solving the First Quadratic Equation

Simplifying the Roots of Equation 1

Resolving the Absolute Value for $tan θ$

Identifying the Larger Root $α_{1}$

Solving the Second Quadratic Equation

Simplifying the Roots of Equation 2

Resolving the Absolute Value for $sec θ$

Identifying the Smaller Root $β_{2}$

Calculating the Sum $α_{1} + β_{2}$

Final Simplification

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The First Quadratic

The Second Quadratic

The Grand Finale

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Visualized Solution

Locating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ on the Unit Circle

Signs of Trigonometric Functions in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q4

Solving the First Quadratic Equation

Simplifying the Roots of Equation 1

Resolving the Absolute Value for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tanθ

Identifying the Larger Root .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α1​