Question

For $0<\theta <\frac{\pi }{2},$ the solution $\left(s\right)$ of $\sum _{m-1}^6$ cosec $(0+\frac{(m-1)\pi }{4})$ cosec $(0+\frac{m\pi }{4})=4\sqrt{2}$ is (are)

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{12}$
• D. $\frac{5\pi }{12}$

Answer 1

Answer: (C), (D)

$\frac{\pi }{12}$
$\frac{5\pi }{12}$

$\sqrt{2}\sum _{m=1}^6\frac{\sin [(0+\frac{m\pi }{4})+(0+\frac{m\pi }{4}-\frac{\pi }{4})]}{\sin (0+\frac{m\pi }{4}-\frac{\pi }{4})}\sin (0+\frac{m\pi }{4})=4\sqrt{2}$
$\Rightarrow \sum _{m=1}^6[\cot (0+\frac{(m-1)\pi }{4})-\cot (0+\frac{m\pi }{4})]=4$
$\Rightarrow \cot 0+\tan 0=4\Rightarrow \tan 0=2\pm \sqrt{3}$
$\Rightarrow 0=\frac{\pi }{12}$ or $\frac{5\pi }{12}$