Question

$\cot \frac{\pi }{20}\cot \frac{3\pi }{20}\cot \frac{7\pi }{20}\cot \frac{9\pi }{20}\cot \frac{15\pi }{20}=$

• A. $1$
• B. $-1$
• C. $\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

The correct option is B $-1$

we know that , $\cot (a+b)=\frac{\cot a\cot b-1}{\cot a+\cot b}\Rightarrow \cot a\cot b=\cot (a+b)[\cot a+\cot b]+1$

So, we can write $\cot \left(\frac{3\pi }{20}\right)\cot \left(\frac{7\pi }{20}\right)=\cot (\frac{3\pi }{20}+\frac{7\pi }{20})\left[\cot \right(\frac{3\pi }{20})+(\frac{7\pi }{20}\left)\right]+1\Rightarrow \cot \left(\frac{3\pi }{20}\right)\cot \left(\frac{7\pi }{20}\right)=\cot \frac{\pi }{2}\left[\cot \right(\frac{3\pi }{20})+(\frac{7\pi }{20}\left)\right]+1\Rightarrow \cot \left(\frac{3\pi }{20}\right)\cot \left(\frac{7\pi }{20}\right)=0+1=1$

Similarly, $\cot \left(\frac{\pi }{20}\right)\cot \left(\frac{9\pi }{20}\right)=1$

and $\cot \left(\frac{15\pi }{20}\right)=\cot \frac{3\pi }{4}=-1$

Now, $\cos \frac{\pi }{7}\cos \frac{2\pi }{7}\cos \frac{4\pi }{7}\cos \frac{5\pi }{7}\cos \frac{6\pi }{7}=-1\ast 1\ast 1\ast =-1$

Therefore, Answer is $-1$