Question

Prove that $ \tan 20^\circ \tan 40^\circ \tan 80^\circ = \tan 60^\circ $

Answer 1

Hint: First of all we will take the left hand side of the equation and will prove its value equal to the right hand side of the equation. Convert the given tangent degrees in the form of the sine upon the cosine angles and then apply different identities for sine into sine and cosine into cosine and then will simplify for the resultant value which should be equal to the right hand side of the equation.

Complete step by step solution:
Given,
LHS $ = \tan 20^\circ \tan 40^\circ \tan 80^\circ $
Re-place tangent as the ratio of sine upon the cosine angles.
LHS $ = \dfrac{{\sin 20^\circ }}{{\cos 20^\circ }}\dfrac{{\sin 40^\circ }}{{\cos 40^\circ }}\dfrac{{\sin 80^\circ }}{{\cos 80^\circ }} $
Re-arrange the above expression –
LHS\[ = \dfrac{{\sin 20^\circ }}{{\cos 20^\circ }}\dfrac{{\sin 80^\circ }}{{\cos 80^\circ }}\dfrac{{\sin 40^\circ }}{{\cos 40^\circ }}\]
Apply the identity for the product of two sine angles and the product of two cosine angles which is expressed as –
 $ \cos \alpha \cos \beta = \dfrac{1}{2}[\cos (\alpha + \beta ) + \cos (\alpha - \beta )] $ and $ \sin \alpha \sin \beta = \dfrac{1}{2}[\cos (\alpha - \beta ) - \cos (\alpha + \beta )] $
LHS $ = \dfrac{{\dfrac{1}{2}[\cos (20^\circ - 80^\circ ) - \cos (20^\circ + 80^\circ )]\sin 40^\circ }}{{\dfrac{1}{2}[\cos (20^\circ - 80^\circ ) + \cos (20^\circ + 80^\circ )]\cos 40^\circ }} $
Simplify the above expression finding the sum and difference of the angles, also common factors from the numerator and the denominator cancel each other.
LHS $ = \dfrac{{[\cos ( - 60^\circ ) - \cos (100^\circ )]\sin 40^\circ }}{{[\cos ( - 60^\circ ) + \cos (100^\circ )]\cos 40^\circ }} $
Cosine is the even function, so $ \cos ( - \theta ) = \cos \theta $ and apply this concept in the above expression
LHS\[ = \dfrac{{[\cos 60^\circ - \cos 100^\circ ]\sin 40^\circ }}{{[\cos 60^\circ + \cos 100^\circ ]\cos 40^\circ }}\]
Place the value for $ \cos 60^\circ = \dfrac{1}{2} $ in the above expression –
LHS \[ = \dfrac{{[\dfrac{1}{2} - \cos 100^\circ ]\sin 40^\circ }}{{[\dfrac{1}{2} + \cos 100^\circ ]\cos 40^\circ }}\]
Take LCM (least common multiple) in the above expression
Common factor from the numerator and the denominator cancels each other
LHS \[ = \dfrac{{[1 - 2\cos 100^\circ ]\sin 40^\circ }}{{[1 + 2\cos 100^\circ ]\cos 40^\circ }}\]
Multiply the term outside the bracket with the terms inside the bracket -
LHS\[ = \dfrac{{\sin 40^\circ - 2\cos 100^\circ \sin 40^\circ }}{{\cos 40^\circ + 2\cos 100^\circ \cos 40^\circ }}\]
Apply identity - $ 2\cos \alpha \sin \beta = [\sin (\alpha + \beta ) - \sin (\alpha - \beta )] $ and $ 2\cos \alpha \cos \beta = [\cos (\alpha + \beta ) + \cos (\alpha - \beta )] $
LHS \[ = \dfrac{{\sin 40^\circ - [\sin (100^\circ + 40^\circ ) - \sin (100^\circ - 40^\circ )]}}{{\cos 40^\circ + [\cos (100^\circ + 40^\circ ) + \cos (100^\circ - 40^\circ )]}}\]
Simplify the above expression finding the sum or difference of the angles –
LHS \[ = \dfrac{{\sin 40^\circ - [\sin 140^\circ - \sin 60^\circ ]}}{{\cos 40^\circ + [\cos 140^\circ + \cos 60^\circ ]}}\]
Open the brackets and use the identity for $ \sin 140^\circ = \sin (180^\circ - 40^\circ ) $ in the above expression –
LHS \[ = \dfrac{{\sin 40^\circ - \sin (180^\circ - 40^\circ ) + \sin 60^\circ }}{{\cos 40^\circ + \cos (180^\circ - 40^\circ ) + \cos 60^\circ }}\]
By identity sine is positive in second quadrant and cosine is negative in second quadrant –
LHS \[ = \dfrac{{\sin 40^\circ - \sin 40^\circ + \sin 60^\circ }}{{\cos 40^\circ - \cos 40^\circ + \cos 60^\circ }}\]
Like terms with the same value and opposite sign cancel each other.
LHS\[ = \dfrac{{\sin 60^\circ }}{{\cos 60^\circ }}\]
The above expression can be written as tangent angle
LHS $ = \tan 60^\circ $
 $ LHS = RHS $

Note: Remember different identities to solve this sum. Be careful about the sign convention and the even and odd trigonometric functions. Remember the All STC rule, that is also known as ASTC rule in the geometry which states that all the trigonometric ratios in the first quadrant ( $ 0^\circ \;{\text{to 90}}^\circ $ ) are positive, sine and cosec are positive in the second quadrant ( $ 90^\circ {\text{ to 180}}^\circ $ ), tan and cot are positive in the third quadrant ( $ 180^\circ \;{\text{to 270}}^\circ $ ) and sin and cosec are positive in the fourth quadrant ( $ 270^\circ {\text{ to 360}}^\circ $ ).