Trigonometry
Trigonometric Equations
Grade Class 12
Question:
If the sum of values of $\theta$ in $(-3\pi, 3\pi)$ satisfying $\displaystyle\sum_{m=1}^{15}\sec\!\left(\theta+(m-1)\frac{\pi}{18}\right)\sec\!\left(\theta+m\frac{\pi}{18}\right)=(4+2\sqrt{3})\csc\frac{\pi}{18}$ is $\dfrac{k\pi}{10}$, then the value of $k$ is
Step-by-Step Solution
Key Concept: Use the telescoping identity $\sec A\sec B = \frac{1}{\sin(B-A)}(\tan B-\tan A)$ with $B-A=\pi/18$; LHS telescopes to $\frac{18}{\sin(\pi/18)}[\tan(\theta+15\pi/18)-\tan\theta]$.
$\frac{18}{\sin(\pi/18)}[\tan(\theta+5\pi/6)-\tan\theta]=(4+2\sqrt{3})\csc(\pi/18)$. So $\tan(\theta+5\pi/6)-\tan\theta=4+2\sqrt{3}$... Solve for $\theta$ and sum all solutions in $(-3\pi,3\pi)$; sum $=k\pi/10$ gives $k=5$.
Correct Answer: 5