Question

The positive integer value of $n>3$ satisfying the equation $\frac{1}{\sin \left(\frac{\pi }{n}\right)}=\frac{1}{\sin \left(\frac{2\pi }{n}\right)}+\frac{1}{\sin \left(\frac{3\pi }{n}\right)}$ is ..

• A. $n=6$
• B. $n=7$
• C. $n=8$
• D. $n=9$

Answer 1

The correct option is B $n=7$

$\frac{1}{\sin \left(\frac{\pi }{n}\right)}=\frac{1}{\sin \left(\frac{2\pi }{n}\right)}+\frac{1}{\sin \left(\frac{3\pi }{n}\right)}$ is

\$\Rightarrow \frac{1}{\sin \left(\frac{\pi }{n}\right)}=\frac{1}{\sin \left(\frac{3\pi }{n}\right)}+\frac{1}{\sin \left(\frac{2\pi }{n}\right)}$

$\Rightarrow \frac{\sin \left(\frac{3\pi }{n}\right)-\sin \left(\frac{\pi }{n}\right)}{\sin \left(\frac{\pi }{n}\right)\sin \left(\frac{3\pi }{n}\right)}=\frac{1}{sin\left(\frac{2\pi }{n}\right)}$ $\left[\sin \right(A+B)-\sin (A-B)=2\sin A\cos B]$

$=\frac{2\sin \left(\frac{\pi }{n}\right)\cos \left(\frac{2\pi }{n}\right)}{\sin \left(\frac{\pi }{n}\right)\sin \left(\frac{3\pi }{n}\right)}\times \sin \left(\frac{2\pi }{n}\right)=1$

$\Rightarrow \frac{\sin \left(\frac{4\pi }{n}\right)}{\sin \left(\frac{3\pi }{n}\right)}=1$ $[\therefore \sin (\pi -\theta )=\sin \theta ]$

$\Rightarrow \sin \left(\frac{4\pi }{n}\right)=\sin \left(\frac{3\pi }{n}\right)\Rightarrow \sin (\pi -\frac{4\pi }{n})=\sin \left(\frac{3\pi }{n}\right)$

$\Rightarrow \pi -\frac{4\pi }{n}=\frac{3\pi }{n}\Rightarrow \frac{7\pi }{n}=7$

$n=7$ So, option (B) is right.