Why does $\tan(30^{\large\circ})=\frac{\tan(10^{\large\circ})\tan(50^{\large\circ})}{\tan(20^{\large\circ})}$?

Solution 1:

Answer to the Question $$ \begin{align} \frac{\tan\left(10^{\large\circ}\right)\tan\left(50^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)} &=\frac1{\tan\left(20^{\large\circ}\right)} \overbrace{\frac{\tan\left(30^{\large\circ}\right)-\tan\left(20^{\large\circ}\right)}{1+\tan\left(30^{\large\circ}\right)\tan\left(20^{\large\circ}\right)}}^{\tan\left(10^{\large\circ}\right)} \overbrace{\frac{\tan\left(30^{\large\circ}\right)+\tan\left(20^{\large\circ}\right)}{1-\tan\left(30^{\large\circ}\right)\tan\left(20^{\large\circ}\right)}}^{\tan\left(50^{\large\circ}\right)}\tag1\\ &=\frac{\frac13-\tan^2\left(20^{\large\circ}\right)}{\tan\left(20^{\large\circ}\right)-\frac13\tan^3\left(20^{\large\circ}\right)}\tag2\\ &=\frac1{\tan\left(60^{\large\circ}\right)}\tag3\\[6pt] &=\tan\left(30^{\large\circ}\right)\tag4 \end{align} $$ Explanation:
$(1)$: $\tan(x+y)=\frac{\tan(x)+\tan(y)}{1-\tan(x)\tan(y)}$
$(2)$: $\tan\left(30^{\large\circ}\right)=\frac1{\sqrt3}$
$(3)$: $\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$
$(4)$: $\frac1{\tan\left(60^{\large\circ}\right)}=\tan\left(90^{\large\circ}-60^{\large\circ}\right)$

General Identity

The answer above is a special case of the identity $$ \cot(3x)=\frac{\tan\left(\frac\pi6-x\right)\tan\left(\frac\pi6+x\right)}{\tan(x)}\tag5 $$ whose proof mirrors the answer above.

Solution 2:

The task is to show that $\tan20^\circ\tan30^\circ=\tan10^\circ\tan50^\circ$. After multiplication through by the cosines of the angles, the task is converted to one of proving that the following quantity is zero:$$a:=\sin20^\circ\sin30^\circ\cos10^\circ\cos50^\circ-\cos20^\circ\cos30^\circ\sin10^\circ\sin50^\circ.$$ Now, from the standard trigonometric formulas $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $2\sin A\sin B=\cos(A-B)-\cos(A+B)$, we get $$2a=(\cos10^\circ-\cos50^\circ)\cos10^\circ\cos50^\circ-(\cos10^\circ+\cos50^\circ)\sin10^\circ\sin50^\circ\qquad\qquad\qquad\quad$$ $$=(\cos10^\circ\cos50^\circ-\sin10^\circ\sin50^\circ)\cos10^\circ-(\cos10^\circ\cos50^\circ+\sin10^\circ\sin50^\circ)\cos50^\circ$$ $$=\cos60^\circ\cos10^\circ-\cos40^\circ\cos50^\circ$$ $$\qquad\qquad=\tfrac12\cos10^\circ-\tfrac12(\cos10^\circ+\cos90^\circ)$$ $$=0.$$

Solution 3:

\begin{align} \frac{\tan10\tan50}{\tan20} &= \frac{\sin10\sin50\cos20}{\cos10\cos50\sin20} = \frac{\sin50\cos20}{2\cos^210\cos50}\\ &= \frac{2\cos40\cos20}{2\cos50(1+\cos20)} = \frac{\cos20+\frac12}{2\cos50+\cos70+\frac{\sqrt3}2}\\ &= \frac{\cos20+\frac12}{\cos50+\cos10+\frac{\sqrt3}2} = \frac{\cos20+\frac12}{\sqrt3\cos20+\frac{\sqrt3}2}\\ &= \frac1{\sqrt3}=\tan30 \end{align}