Question

$\cot \frac{\pi }{20}\cot \frac{3\pi }{20}\cot \frac{5\pi }{20}\cot \frac{7\pi }{20}\cot \frac{9\pi }{20}$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

We know that $\pi ={180}^{\circ }$

$\frac{\pi }{20}=9^{\circ }$, $\frac{3\pi }{20}={27}^{\circ }$, $\frac{5\pi }{20}={45}^{\circ }$, $\frac{7\pi }{20}={63}^{\circ }$, $\frac{9\pi }{20}={81}^{\circ }$

$\cot \frac{\pi }{20}\cot \frac{3\pi }{20}\cot \frac{5\pi }{20}\cot \frac{7\pi }{20}\cot \frac{9\pi }{20}$

$=\cot 9^{\circ }\cot {27}^{\circ }\cot {45}^{\circ }\cot {63}^{\circ }\cot {81}^{\circ }$

$=\cot 9^{\circ }\cot {27}^{\circ }\times 1\times \cot ({90}^{\circ }-{27}^{\circ })\cot ({90}^{\circ }-9^{\circ })$

$=\cot 9^{\circ }\cot {27}^{\circ }\tan 9^{\circ }\tan {27}^{\circ }$

$=\cot 9^{\circ }\tan 9^{\circ }\cot {27}^{\circ }\tan {27}^{\circ }$

$=1$ since $\tan \theta \cot \theta =1$