Question

The value of ${\cot }^2\frac{\pi }{9}+{\cot }^2\frac{2\pi }{9}+{\cot }^2\frac{4\pi }{9}$, is

• A. $0$
• B. $3$
• C. $9$
• D. $1/3$

Answer 1

The correct option is C $9$

${\cot }^2\frac{\pi }{9}+{\cot }^2\frac{2\pi }{9}+{\cot }^2\frac{4\pi }{9}$

$\Rightarrow {\csc }^2\frac{\pi }{9}+{\csc }^2\frac{2\pi }{9}+{\csc }^2\frac{4\pi }{9}-3$

$\Rightarrow \frac{1}{{\sin }^2\pi /9}+\frac{1}{{\sin }^{2}2\pi /9}+\frac{1}{{\sin }^{2}4\pi /9}-3$

$=\frac{2}{1-\cos 2\pi /9}+\frac{2}{1-\cos 4\pi /9}+\frac{2}{1-\sin 8\pi /9}-3$

$=\frac{2}{1+\cos 7\pi /9}+\frac{2}{1-\cos 5\pi /9}+\frac{2}{1-\cos \pi /9}-3$

$\Rightarrow 2\left[\frac{(1+\cos 5\pi /9)(1+\cos \pi /9)+(1+\cos 7\pi /9)(1+\cos \pi /9)+(1+\cos 7\pi /9)(1+\cos 5\pi /9)}{(1+\cos 7\pi /9)(1+\cos 5\pi /9)(1+\cos 9\pi /9)}\right]-3$

$=2\left[\frac{3+2\cos \frac{\pi }{9}+2\cos \frac{5\pi }{9}+2\cos \frac{7\pi }{9}+\cos \frac{5\pi }{9}\cos \frac{\pi }{9}+\cos \frac{7\pi }{9}\cos \frac{\pi }{9}+\cos \frac{9\pi }{9}\cos \frac{5\pi }{9}}{\left(\right)\left(\right)\left(\right)}\right]-3$

$=\left[\frac{3+2\cos \frac{\pi }{9}+2\cos \frac{5\pi }{9}+2\cos \frac{7\pi }{9}+\frac{1}{2}[\cos \frac{6\pi }{9}+\cos \frac{4\pi }{9}+\cos \frac{8\pi }{9}+\cos \frac{6\pi }{9}+\cos \frac{12\pi }{9}+\cos \frac{2\pi }{9}]}{\left(\right)\left(\right)\left(\right)}\right]-3$

$=2\left[\frac{3+2\cos \frac{\pi }{9}+2\cos \frac{5\pi }{9}+2\cos \frac{7\pi }{9}+\frac{1}{2}[-\cos \frac{3\pi }{9}-\cos \frac{5\pi }{9}-\cos \frac{\pi }{9}-\cos \frac{3\pi }{9}-\cos \frac{3\pi }{9}-\cos \frac{7\pi }{9}]}{\left(\right)\left(\right)\left(\right)}\right]-3$