Comment
Share
Q)

# The positive integer value of $n>3$ satisfying the equation $\large\frac{1}{\sin\big(\Large\frac{\pi}{n}\big)}=\frac{1}{\sin\big(\Large\frac{2\pi}{n}\big)}+\frac{1}{\sin\big(\Large\frac{3\pi}{n}\big)}$ is

$(a)n=6\qquad(b)\;n=7\qquad(c)\;n=8\qquad(d)\;n=9$

• $sin x-siny=2cos\large\frac{x+y}{2}.$$sin\large\frac{x-y}{2} • 2sinx.cosx=sin2x Given : \large\frac{1}{\sin\big(\Large\frac{\pi}{n}\big)}=\frac{1}{\sin\big(\Large\frac{2\pi}{n}\big)}+\frac{1}{\sin\big(\Large\frac{3\pi}{n}\big)} \Rightarrow \large\frac{1}{\sin\pi/n}-\frac{1}{\sin \large\frac{3\pi}{n}}=\frac{1}{\sin\Large\frac{2\pi}{n}} \Rightarrow \large\frac{\sin 3\pi/n-\sin\pi/n}{\sin\pi/n.\sin 3\pi/n}$$=\large\frac{1}{sin\large\frac{2\pi}{n}}$
$\Rightarrow\:\large\frac{2cos 2\pi/n.sin \pi/n}{sin\pi/n.sin3\pi/n}$$=\Rightarrow \large\frac{1}{\sin\large\frac{2\pi}{n}} \Rightarrow 2\cos\big(\large\frac{2\pi}{n}\big).$$\sin\large\frac{2\pi}{n}$$=sin\frac{3\pi}{n} \Rightarrow 2\sin \Large\frac{2\pi}{n}.$$\cos\Large\frac{2\pi}{n}=$$\sin\large\frac{3\pi}{n} \Rightarrow \sin\large\frac{4\pi}{n}-$$\sin\large\frac{3\pi}{n}$
$\Rightarrow \large\frac{4\pi}{n}=$$\pi-\large\frac{3\pi}{n} \Rightarrow \large\frac{7\pi}{n}=$$\pi$
$n=7$