Question

Let \(S=\left\{0∈(0,\frac{π}{2}) : \sum^{9}_{m=1} \sec(θ+(m-1)\frac{π}{6})\sec(θ+\frac{mπ}{6}) = -\frac{8}{\sqrt3}\right\}\)
Then,

• A. \(S = \left\{\frac{π}{12}\right\}\)
• B. \(S = \left\{\frac{2π}{3}\right\}\)
• C. \(∑_{θ∈S}θ = \frac{π}{2}\)
• D. \(∑_{θ∈S}θ = \frac{3π}{4}\)

Answer 1

The Correct Option is C

To solve the problem, we are tasked with evaluating the set \( S \) given by:

\( S = \left\{ \theta \in \left(0, \frac{\pi}{2}\right) : \sum_{m=1}^{9} \sec\left(\theta + (m-1)\frac{\pi}{6}\right)\sec\left(\theta + \frac{m\pi}{6}\right) = -\frac{8}{\sqrt{3}} \right\} \)

We need to find the value of \( ∑_{\theta∈S}θ \).

First, consider each term in the summation:

\(\sec(x + y) = \frac{1}{\cos(x + y)}\), and we use the angle addition formula:
\(\cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b)\).

This problem simplifies to solving given constraint over possible values of \( \theta \). Due to the symmetry and nature of trigonometric functions, \( \theta \) must be resolved in one cycle \( \left(0, \frac{\pi}{2}\right) \).

We evaluate possibilities observing that periodic properties of trigonometric functions result in cancellations or augmentations over multiples of \( \frac{\pi}{6} \), producing repetitive behavior.

Applying the given equation constraint on respective intervals yields:

The correct summed value is when the characteristic balance of the sum relates as requested i.e. \( ∑_{\theta∈S}\theta = \frac{\pi}{2} \).

Answer 2

\[ S=\Big\{\theta\in(0,\tfrac{\pi}{2}) : \sum_{m=1}^{9} \sec\!\big(\theta+(m-1)\tfrac{\pi}{6}\big)\, \sec\!\big(\theta+m\tfrac{\pi}{6}\big) \;=\; -\tfrac{8}{\sqrt{3}}\Big\}. \]

Step 1: Use the identity \( \tan b - \tan a = \dfrac{\sin(b-a)}{\cos a \cos b} \)

\[ \sec\alpha\,\sec\beta \;=\; \frac{1}{\cos\alpha\cos\beta} \;=\; \frac{\tan\beta-\tan\alpha}{\sin(\beta-\alpha)}. \] Here, for each \(m\), \(\;\alpha=\theta+(m-1)\tfrac{\pi}{6}\), \(\;\beta=\theta+m\tfrac{\pi}{6}\), so \(\beta-\alpha=\tfrac{\pi}{6}\) and \(\sin(\tfrac{\pi}{6})=\tfrac{1}{2}\). Hence \[ \sec\!\big(\theta+(m-1)\tfrac{\pi}{6}\big)\, \sec\!\big(\theta+m\tfrac{\pi}{6}\big) \;=\; 2\Big[\tan\!\big(\theta+m\tfrac{\pi}{6}\big)-\tan\!\big(\theta+(m-1)\tfrac{\pi}{6}\big)\Big]. \]

Step 2: Telescoping the sum

\[ \sum_{m=1}^{9}\sec(\cdots)\sec(\cdots) \;=\; 2\sum_{m=1}^{9}\Big[\tan\!\big(\theta+m\tfrac{\pi}{6}\big)-\tan\!\big(\theta+(m-1)\tfrac{\pi}{6}\big)\Big] \] \[ =\; 2\Big[\tan\!\big(\theta+9\tfrac{\pi}{6}\big)-\tan\theta\Big] \;=\; 2\big[\tan(\theta+\tfrac{3\pi}{2})-\tan\theta\big]. \] Using \(\tan(\theta+\tfrac{\pi}{2})=-\cot\theta\) and \(\tan(\theta+\pi)=\tan\theta\), \(\tan(\theta+\tfrac{3\pi}{2})=\tan\big((\theta+\pi)+\tfrac{\pi}{2}\big)=-\cot\theta\). Therefore \[ \sum_{m=1}^{9}\sec(\cdots)\sec(\cdots)=2[-\cot\theta-\tan\theta]=-2(\cot\theta+\tan\theta). \]

Step 3: Solve for \( \theta \)

Given the sum equals \(-\dfrac{8}{\sqrt{3}}\), we get \[ -2(\cot\theta+\tan\theta)=-\frac{8}{\sqrt{3}} \;\;\Longrightarrow\;\; \cot\theta+\tan\theta=\frac{4}{\sqrt{3}}. \] But \(\;\cot\theta+\tan\theta=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} =\dfrac{1}{\sin\theta\cos\theta}\). Hence \[ \frac{1}{\sin\theta\cos\theta}=\frac{4}{\sqrt{3}} \;\;\Longrightarrow\;\; \sin\theta\cos\theta=\frac{\sqrt{3}}{4} \;\;\Longrightarrow\;\; \sin 2\theta=2\sin\theta\cos\theta=\frac{\sqrt{3}}{2}. \] With \(\theta\in(0,\tfrac{\pi}{2})\), the solutions are \[ 2\theta=\frac{\pi}{3},\;\frac{2\pi}{3} \;\;\Longrightarrow\;\; \theta=\frac{\pi}{6},\;\frac{\pi}{3}. \] Thus \(S=\Big\{\tfrac{\pi}{6},\,\tfrac{\pi}{3}\Big\}\).

\[ \sum_{\theta\in S}\theta \;=\; \frac{\pi}{6}+\frac{\pi}{3} \;=\; \boxed{\frac{\pi}{2}}. \]