Question

tan 2π/18 tan 4π/18 tan 8π/18 = √k find the value of k ​

Answer 1

Step-by-step explanation:

We have,

$\tt{tan \bigg( \dfrac{2\pi}{18} \bigg) tan \bigg( \dfrac{4\pi}{18} \bigg) tan \bigg( \dfrac{8\pi}{18} \bigg) }$

$= \sf{tan \bigg( \dfrac{\pi}{9} \bigg) tan \bigg( \dfrac{2\pi}{9} \bigg) tan \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ \dfrac{sin \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ cos \bigg( \dfrac{\pi}{9} \bigg) cos\bigg( \dfrac{2\pi}{9} \bigg) cos\bigg( \dfrac{4\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 2 sin \bigg( \dfrac{\pi}{9} \bigg) \cdot \: sin \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ 2 sin \bigg( \dfrac{\pi}{9} \bigg) cos \bigg( \dfrac{\pi}{9} \bigg) cos\bigg( \dfrac{2\pi}{9} \bigg) cos\bigg( \dfrac{4\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 2 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ sin \bigg( \dfrac{2\pi}{9} \bigg) cos\bigg( \dfrac{2\pi}{9} \bigg) cos\bigg( \dfrac{4\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 4 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ 2sin \bigg( \dfrac{2\pi}{9} \bigg) cos\bigg( \dfrac{2\pi}{9} \bigg) cos\bigg( \dfrac{4\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 4 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ sin \bigg( \dfrac{4\pi}{9} \bigg) cos\bigg( \dfrac{4\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 8 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ 2sin \bigg( \dfrac{4\pi}{9} \bigg) cos\bigg( \dfrac{4\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 8 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ sin \bigg( \dfrac{8\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 8 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ sin \bigg( \pi - \dfrac{\pi}{9} \bigg) } }$

$= \sf{ \dfrac{ 8 sin ^{2} \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg)}{ sin \bigg( \dfrac{\pi}{9} \bigg) } }$

$= \sf{ 8 sin \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 4 \cdot2sin \bigg( \dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{2\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 4 \cdot \bigg \{cos \bigg( \dfrac{2\pi}{9} - \dfrac{\pi}{9} \bigg) - cos \bigg( \dfrac{2\pi}{9} + \dfrac{\pi}{9} \bigg)\bigg \} sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 4 \cdot \bigg \{cos \bigg(\dfrac{\pi}{9} \bigg) - cos \bigg( \dfrac{3\pi}{9} \bigg)\bigg \} sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 4 \cdot \bigg \{cos \bigg(\dfrac{\pi}{9} \bigg) - cos \bigg( \dfrac{\pi}{3} \bigg)\bigg \} sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 4 \cdot \bigg \{cos \bigg(\dfrac{\pi}{9} \bigg) - \dfrac{1}{2} \bigg \} sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 4 cos \bigg(\dfrac{\pi}{9} \bigg) sin \bigg( \dfrac{4\pi}{9} \bigg) - 2 sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 2 \cdot2 sin \bigg( \dfrac{4\pi}{9} \bigg) cos \bigg(\dfrac{\pi}{9} \bigg) - 2 sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 2 \cdot \bigg \{ sin \bigg( \dfrac{4\pi}{9} + \dfrac{\pi}{9} \bigg) + sin \bigg( \dfrac{4\pi}{9} - \dfrac{\pi}{9} \bigg) \bigg \} - 2 sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 2 sin \bigg( \dfrac{5\pi}{9} \bigg) + 2sin \bigg( \dfrac{3\pi}{9} \bigg) - 2 sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 2 sin \bigg(\pi - \dfrac{4\pi}{9} \bigg) + 2sin \bigg( \dfrac{\pi}{3} \bigg) - 2 sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ 2 sin \bigg( \dfrac{4\pi}{9} \bigg) + 2 \times \dfrac{ \sqrt{3} }{2} - 2 sin \bigg( \dfrac{4\pi}{9} \bigg) }$

$= \sf{ \sqrt{3} }$

So, the required value of k=3