Prove That:Tan 20° Tan 40° Tan 60° Tan 80° = 3 - Mathematics

Sum

Prove that:
tan 20° tan 40° tan 60° tan 80° = 3

Solution

LHS = tan 20° tan 40° tan 60° tan 80°
$= \tan 60^\circ \frac{\sin 20^\circ \sin 40^\circ \sin 80^\circ} {\cos 20^\circ \cos 40^\circ \cos 80^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{2}\left[ 2\sin 20^\circ \sin 40^\circ \right]\sin 80^\circ }{\frac{1}{2}\left[ 2\cos 20^\circ \cos 40^\circ \right]\cos 80^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{2}\left[ \cos \left( 20^\circ - 40^\circ \right) - \cos \left( 20^\circ + 40^\circ \right) \right] \sin 80^\circ}{\frac{1}{2}\left[ \cos \left( 20^\circ + 40^\circ \right) + \cos\left( 20^\circ - 40^\circ \right) \right] \cos 80^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{2}\left[ \cos \left( - 20^\circ \right) - \cos 60^\circ \right] \sin 80^\circ}{\frac{1}{2}\left[ \cos 60^\circ + \cos\left( - 20^\circ \right) \right] \cos 80^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{2}\sin 80^\circ\left[ \cos 20^\circ - \frac{1}{2} \right]}{\frac{1}{2}\cos 80^\circ\left[ \frac{1}{2} + \cos 20^\circ \right]}$
$= \sqrt{3} \times \frac{\frac{1}{2}\sin 80^\circ \cos 20^\circ - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{2}\cos 80^\circ \cos20^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{2}\sin \left( 90^\circ - 10^\circ \right) \cos 20^\circ - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{2}\cos 80^\circ \cos 20^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{2}\cos 10^\circ \cos 20^\circ - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{2}\cos 80^\circ \cos 20^\circ}$
$= \sqrt{3} \times \frac{\frac{1}{4}\left[ 2\cos 10^\circ \cos 20^\circ \right] - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ 2\cos 80^\circ \cos 20^\circ \right]}$
$= \sqrt{3} \times \frac{\frac{1}{4}\left[ \cos \left( 10^\circ + 20^\circ \right) + \cos \left( 10^\circ - 20^\circ \right) \right] - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ \cos \left( 80^\circ + 20^\circ \right) + \cos \left( 80^\circ - 20^\circ \right) \right]}$
$= \sqrt{3} \times \frac{\frac{1}{4}\left[ \cos 30^\circ + \cos \left( - 10^\circ \right) \right] - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ \cos 100^\circ + \cos 60^\circ \right]}$
$= \sqrt{3} \times \frac{\frac{1}{4}\left[ \cos 30^\circ + \cos \left( 90^\circ - 80^\circ \right) \right] - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ + \frac{1}{4}\left[ \cos \left( 180^\circ - 80^\circ \right) + \frac{1}{2} \right]}$
$= \sqrt{3} \times \frac{\frac{\sqrt{3}}{8} + \frac{1}{4}\sin 80^\circ - \frac{1}{4}\sin 80^\circ}{\frac{1}{4}\cos 80^\circ - \frac{1}{4}\cos 80^\circ + \frac{1}{8}}\left[ \cos \left( 90^\circ - 80^\circ \right) = \sin 80^\circ, and \cos\left( 180^\circ - 80^\circ \right) = - \cos\left( 80^\circ \right) \right]$
$= \sqrt{3} \times \frac{\frac{\sqrt{3}}{8}}{\frac{1}{8}}$
$= 3 = RHS$

Concept: Transformation Formulae
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RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Exercise 8.1 | Q 5.5 | Page 7