Question

Prove that $\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = 1$.

Answer 1

We can rearrange the terms in the product. Express $\tan$ in terms of sine and cosine. Then we can apply the result of addition and subtraction of cosine terms. Now substitute the known trigonometric values. Simplifying we get the left hand side equal to the right hand side.

Formula used:
We have the trigonometric relations:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\cos A - \cos B = - 2\sin (\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2}) - - - (i)$
$\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2}) - - - (ii)$
$\sin ( - \theta ) = - \sin \theta$
$\cos ( - \theta ) = \cos \theta$
Also we have,
$(a + b)(a - b) = {a^2} - {b^2}$

Complete step-by-step answer:
Given the expression,
$\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ$
We can rearrange the terms.
$\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = (\tan 66^\circ \tan 6^\circ )(\tan 78^\circ \tan 42^\circ )$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
Using this we get,
$\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = (\dfrac{{\sin 66^\circ }}{{\cos 66^\circ }}\dfrac{{\sin 6^\circ }}{{\cos 6^\circ }})(\dfrac{{\sin 78^\circ }}{{\cos 78^\circ }}\dfrac{{\sin 42^\circ }}{{\cos 42^\circ }})$
Multiplying and dividing by $2$ we get,
$\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = (\dfrac{{2\sin 66^\circ }}{{2\cos 66^\circ }}\dfrac{{\sin 6^\circ }}{{\cos 6^\circ }})(\dfrac{{2\sin 78^\circ }}{{2\cos 78^\circ }}\dfrac{{\sin 42^\circ }}{{\cos 42^\circ }}) - - - *$

We have the trigonometric relations:
$\cos A - \cos B = - 2\sin (\dfrac{{A + B}}{2})\sin (\dfrac{{A - B}}{2}) - - - (i)$
$\cos A + \cos B = 2\cos (\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2}) - - - (ii)$
Let $A = 60,B = 72$
We get, $\dfrac{{A + B}}{2} = \dfrac{{60 + 72}}{2} = \dfrac{{132}}{2} = 66$
$\dfrac{{A - B}}{2} = \dfrac{{60 - 72}}{2} = \dfrac{{ - 12}}{2} = - 6$
Substituting in the above result we get,
$\Rightarrow$ $\cos 60 - \cos 72 = - 2\sin 66\sin ( - 6)$
We have $\sin ( - \theta ) = - \sin \theta$
So we get,
$\Rightarrow$ $\cos 60 - \cos 72 = 2\sin 66\sin 6 - - - (iii)$
Also $(ii)$ gives,
$\Rightarrow$ $\cos 60 + \cos 72 = 2\cos 66\cos ( - 6)$
And we have $\cos ( - \theta ) = \cos \theta$
So we get,
$\cos 60 + \cos 72 = 2\cos 66\cos 6 - - - (iv)$
Similarly let $A = 36,B = 120$
We get, $\dfrac{{A + B}}{2} = \dfrac{{36 + 120}}{2} = \dfrac{{156}}{2} = 78$
$\Rightarrow$ $\dfrac{{A - B}}{2} = \dfrac{{36 - 120}}{2} = \dfrac{{ - 84}}{2} = - 42$
Substituting in the above result we get,
$\Rightarrow$ $\cos 36 - \cos 120 = - 2\sin 78\sin ( - 42)$
We have $\sin ( - \theta ) = - \sin \theta$
So we get,
$\Rightarrow$ $\cos 36 - \cos 120 = 2\sin 78\sin 42 - - - (v)$
Also $(ii)$ gives,
$\Rightarrow$ $\cos 36 + \cos 120 = 2\cos 78\cos ( - 42)$
And we have $\cos ( - \theta ) = \cos \theta$
So we get,
$\Rightarrow$ $\cos 36 + \cos 120 = 2\cos 78\cos 42 - - - (vi)$

Substituting using equations $(iii),(iv),(v)$ and $(vi)$ in equation $*$ we get,
$\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = (\dfrac{{\cos 60 - \cos 72}}{{\cos 60 + \cos 72}})(\dfrac{{\cos 36 - \cos 120}}{{\cos 36 + \cos 120}}) - - - **$
We know $\cos 60 = \dfrac{1}{2}$ and $\cos 120 = \cos (90 + 30) = - \sin 30 = - \dfrac{1}{2}$
Also $72 = 90 - 18$
This gives $\cos 72 = \cos (90 - 18) = = \sin 18$
Substituting these we get,
$\Rightarrow$ $\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = (\dfrac{{\dfrac{1}{2} - \sin 18}}{{\dfrac{1}{2} + \sin 18}}) \times (\dfrac{{\cos 36 - ( - \dfrac{1}{2})}}{{\cos 36 + ( - \dfrac{1}{2})}})$
Simplifying we get,
$\Rightarrow$ $\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = (\dfrac{{\dfrac{1}{2} - \sin 18}}{{\dfrac{1}{2} + \sin 18}}) \times (\dfrac{{\cos 36 + \dfrac{1}{2}}}{{\cos 36 - \dfrac{1}{2}}})$
We have $\sin 18 = \dfrac{{\sqrt 5 - 1}}{4}$ and $\cos 36 = \dfrac{{\sqrt 5 + 1}}{4}$.
Substituting we get,
$\Rightarrow$ $\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \dfrac{{\dfrac{1}{2} - \dfrac{{\sqrt 5 - 1}}{4}}}{{\dfrac{1}{2} + \dfrac{{\sqrt 5 - 1}}{4}}} \times \dfrac{{\dfrac{{\sqrt 5 + 1}}{4} + \dfrac{1}{2}}}{{\dfrac{{\sqrt 5 + 1}}{4} - \dfrac{1}{2}}}$
Simplifying we get,
$\Rightarrow$ $\tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \dfrac{{\dfrac{{2 - \sqrt 5 + 1}}{4}}}{{\dfrac{{2 + \sqrt 5 - 1}}{4}}} \times \dfrac{{\dfrac{{\sqrt 5 + 1 + 2}}{4}}}{{\dfrac{{\sqrt 5 + 1 - 2}}{4}}}$
$\Rightarrow \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \dfrac{{3 - \sqrt 5 }}{{1 + \sqrt 5 }} \times \dfrac{{3 + \sqrt 5 }}{{\sqrt 5 - 1}}$
We know that
$(a + b)(a - b) = {a^2} - {b^2}$
Using this we get,
$\Rightarrow \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \dfrac{{9 - 5}}{{5 - 1}}$
$\Rightarrow \tan 6^\circ \tan 42^\circ \tan 66^\circ \tan 78^\circ = \dfrac{4}{4} = 1$
Hence we had proved the result.

Note: We can show that $\sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}{\text{ }}$
For, consider $\theta = {18^ \circ }$
$\Rightarrow 5\theta = 18 \times 5 = {90^ \circ }$
We can write it as
$\Rightarrow 2\theta + 3\theta = {90^ \circ }$
$\Rightarrow 2\theta = {90^ \circ } - 3\theta$
Taking sine on both sides we get,
$\Rightarrow \sin 2\theta = \sin ({90^ \circ } - 3\theta )$
We know $\sin ({90^ \circ } - \theta ) = \cos \theta$
$\Rightarrow \sin 2\theta = \cos 3\theta$
Also we have the trigonometric relations,
$\sin 2\theta = 2\sin \theta \cos \theta$
$\cos 3\theta = 4{\cos ^3}\theta - 3\cos \theta$
Using these in the above equation we get,
$\Rightarrow 2\sin \theta \cos \theta = 4{\cos ^3}\theta - 3\cos \theta$
$\Rightarrow 2\sin \theta \cos \theta - 4{\cos ^3}\theta + 3\cos \theta = 0$
Taking $\cos \theta$ common we get,
$\Rightarrow \cos \theta (2\sin \theta - 4{\cos ^2}\theta + 3) = 0$
Product of two terms equal to zero means any of the terms is zero.
Since $\theta = {18^ \circ }$, $\cos \theta \ne 0$.
So we have,
$\Rightarrow 2\sin \theta - 4{\cos ^2}\theta + 3 = 0$
Again, ${\cos ^2}\theta = 1 - {\sin ^2}\theta$
$\Rightarrow 2\sin \theta - 4(1 - {\sin ^2}\theta ) + 3 = 0$
$\Rightarrow 2\sin \theta - 4 + 4{\sin ^2}\theta + 3 = 0$
Simplifying we get,
$\Rightarrow 2\sin \theta + 4{\sin ^2}\theta - 1 = 0$
$\Rightarrow 4{\sin ^2}\theta + 2\sin \theta - 1 = 0$
So we get a quadratic equation in $\sin \theta$,
A quadratic equation in the form $a{x^2} + bx + c = 0$ has the solution, $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Solving using this we get,
$\sin \theta = \dfrac{{ - 2 \pm \sqrt {{2^2} - (4 \times 4 \times - 1)} }}{{2 \times 4}}$
$\Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {4 + 16} }}{8}$
$\Rightarrow \sin \theta = \dfrac{{ - 2 \pm \sqrt {20} }}{8} = \dfrac{{ - 2 \pm 2\sqrt 5 }}{8}$
Dividing numerator and denominator by $2$ we get,
$\Rightarrow \sin \theta = \dfrac{{ - 1 \pm \sqrt 5 }}{4}$
Substituting for $\theta$ we get,
$\Rightarrow \sin {18^ \circ } = \dfrac{{ - 1 \pm \sqrt 5 }}{4}$
Since $\sin {18^ \circ }$ is positive, we have
$\Rightarrow \sin {18^ \circ } = \dfrac{{ - 1 + \sqrt 5 }}{4}$
$\Rightarrow \sin {18^ \circ } = \dfrac{{\sqrt 5 - 1}}{4}$
Similarly we can show that $\cos 36 = \dfrac{{\sqrt 5 + 1}}{4}$