Question

Let $\dfrac{{ - \pi }}{6} < \theta < \dfrac{{ - \pi }}{{12}}$, suppose ${\alpha _1}$ and ${\beta _1}$ are the equation ${x^2} - 2x\sec \theta + 1 = 0$ and ${\alpha _2}$ and ${\beta _2}$ are the roots of the equation ${x^2} + 2x\tan \theta - 1 = 0$. If ${\alpha _1} > {\beta _1}$ and ${\alpha _2} > {\beta _2}$, then find the value of ${\alpha _1} + {\beta _2}$.
(A) $2\left( {\sec \theta - \tan \theta } \right)$
(B) $2\sec \theta$
(C) $- 2\tan \theta$
(D) $0$

Answer 1

Analyse the problem properly before starting a solution. Start with finding the roots of quadratic equations using the Quadratic formula. Now use the interval of angle $\dfrac{{ - \pi }}{6} < \theta < \dfrac{{ - \pi }}{{12}}$ to find the value of $\alpha ,\beta$ according to the relation ${\alpha _1} > {\beta _1}$ and ${\alpha _2} > {\beta _2}$. After finding roots separately, just evaluate ${\alpha _1} + {\beta _2}$.

Complete step-by-step answer:
Firstly, we should analyse the given information in the question. It is given that angle $\theta$ lies in an interval $\left( { - 30^\circ , - 15^\circ } \right)$. And we are given with two quadratic equations ${x^2} - 2x\sec \theta + 1 = 0$ and ${x^2} + 2x\tan \theta - 1 = 0$ with the respective roots as ${\alpha _1}$, ${\beta _1}$ and ${\alpha _2}$, ${\beta _2}$. Also, we know that $\alpha > \beta$ for both the pairs.
We can start by finding the roots of the equation ${x^2} - 2x\sec \theta + 1 = 0$ first. This can be done by using Quadratic formula, that can define for an equation of the form $a{x^2} + bx + c = 0$ as:
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Using the Quadratic formula, we can write the roots of the equation ${x^2} - 2x\sec \theta + 1 = 0$ as:
$\Rightarrow x = \dfrac{{ - \left( { - 2\sec \theta } \right) \pm \sqrt {{{\left( { - 2\sec \theta } \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}$
This can be further simplified as:
$\Rightarrow x = \dfrac{{2\sec \theta \pm \sqrt {4{{\sec }^2}\theta - 4} }}{2} = \dfrac{{2\sec \theta \pm 2\sqrt {{{\sec }^2}\theta - 1} }}{2}$
We can now divide the numerator by $2$, so we get:
$\Rightarrow x = \sec \theta \pm \sqrt {{{\sec }^2}\theta - 1}$
Also, we know that: $1 + {\tan ^2}\theta = {\sec ^2}\theta \Rightarrow {\tan ^2}\theta = {\sec ^2}\theta + 1$ . Substituting that:
$\Rightarrow x = \sec \theta \pm \sqrt {{{\tan }^2}\theta } = \sec \theta \pm \tan \theta$ (1)
Now, let’s consider the equation ${x^2} + 2x\tan \theta - 1 = 0$ and apply the Quadratic equation in this:
$\Rightarrow x = \dfrac{{ - 2\tan \theta \pm \sqrt {{{\left( {2\tan \theta } \right)}^2} - 4 \times 1 \times \left( { - 1} \right)} }}{{2 \times 1}}$
Now, let’s take $4$ out of the radical sign and rewrite it as:
$\Rightarrow x = \dfrac{{ - 2\tan \theta \pm \sqrt {4{{\tan }^2}\theta + 4} }}{2} = \dfrac{{ - 2\tan \theta \pm 2\sqrt {{{\tan }^2}\theta + 1} }}{2} = - \tan \theta \pm \sqrt {{{\tan }^2}\theta + 1}$
Again by using: $1 + {\tan ^2}\theta = {\sec ^2}\theta \Rightarrow {\tan ^2}\theta = {\sec ^2}\theta + 1$, we get:
$\Rightarrow x = - \tan \theta \pm \sqrt {{{\tan }^2}\theta + 1} = - \tan \theta \pm \sec \theta$ (2)
Now, we already know that angle $\theta$ lies in $\left( { - 30^\circ , - 15^\circ } \right)$, therefore both $\sec \theta$ and $\tan \theta$ are negative in this interval. Since $\theta$ lies in fourth quadrant and Secant and tangent functions have negative values for fourth quadrant angles.
$\Rightarrow \sec \theta < 0,\tan \theta < 0$ for all $\theta$in the interval $\left( { - 30^\circ , - 15^\circ } \right)$
So, from relation (1) and (2), we can conclude from ${\alpha _1} > {\beta _1}$ and ${\alpha _2} > {\beta _2}$ that:
${\alpha _1} = \sec \theta - \tan \theta$
${\beta _1} = \sec \theta + \tan \theta$
${\alpha _2} = \sec \theta - \tan \theta$
${\beta _2} = - \sec \theta - \tan \theta$
Therefore, the required value: ${\alpha _1} + {\beta _2} = \sec \theta - \tan \theta - sec\theta - \tan \theta = - 2\tan \theta$

Hence, the option (C) is the correct option.

Note: Solve the quadratic formula carefully. The given angle interval $\dfrac{{ - \pi }}{6} < \theta < \dfrac{{ - \pi }}{{12}}$ can be written as $\left( { - 30^\circ , - 15^\circ } \right)$ because $\pi$ radians of angle is equal to $180^\circ$ angle. So $\dfrac{{ - \pi }}{6} = 30^\circ$ can be a different way of writing the angles. While figuring out the value of ${\alpha _1},{\beta _1}$ using the solution of the respective equation carefully. The sign of secant and tangent function will determine the larger root.