I came across another trigonometry problem and it is as following:
Prove that $\tan55^\circ\cdot\tan65^\circ\cdot\tan75^\circ=\tan85^\circ$.
I know that we can write $\tan 75^\circ=\cot{3\cdot5^\circ}=\frac{1-3\tan^2{5^\circ}}{3\tan 5^\circ-\tan^3{5^\circ}}$ and also $\tan 85^\circ=\frac{1}{\tan 5^\circ}$ so now we should prove $\tan 55^\circ\cdot\tan 65^\circ=\frac{3-\tan^2{5^\circ}}{1-3\tan^2{5^\circ}}$.
I don't know how to do that from here. Any ideas?


You've done well till now! I'll give you a hint to do the remaining. Note that:

$$ \begin{align} \tan 55^\circ \cdot \tan 65^\circ &= \tan(60^\circ – 5^\circ )\cdot \tan( 60^\circ + 5^\circ ) \\ &= \dfrac{ \tan 60^\circ – \tan 5^\circ }{ 1 + \tan 60^\circ \cdot \tan 5^\circ}\dfrac{ \tan 60^\circ +\tan 5^\circ }{ 1 - \tan 60^\circ \cdot \tan 5^\circ } \\ & = \dfrac{\left(\dfrac{ \sqrt{3} }{2} + \tan 5^\circ\right)\left(\dfrac{\sqrt{3}}{2} - \tan 5^\circ\right)}{1-\tan^260^\circ.\tan^25^\circ} \end{align} $$

With using $x^2-y^2$ = $(x-y)(x+y)$ there's only a little left :) Hope this could help.

  • $\begingroup$ @user148256 You're welcome :) $\endgroup$ – CODE May 6 '14 at 17:42

Consider the identity $\tan x\cdot\tan(60^\circ-x)\cdot\tan(60^\circ+x)=\tan 3x$. Now, let $x=5^\circ$, we have $$ \tan 5^\circ\cdot\tan55^\circ\cdot\tan65^\circ=\tan 15^\circ.\tag1 $$ Divide $(1)$ by $\tan 5^\circ$ yield $$ \begin{align} \tan55^\circ\cdot\tan65^\circ&=\frac{\tan 15^\circ}{\tan5^\circ}\\ &=\tan 15^\circ\cdot\cot5^\circ\\ &=\tan 15^\circ\cdot\tan(90^\circ-5^\circ)\\ &=\tan 15^\circ\cdot\tan85^\circ.\tag2 \end{align} $$ Divide $(2)$ by $\tan 15^\circ$ yield $$ \begin{align} \frac{\tan55^\circ\cdot\tan65^\circ}{\tan 15^\circ}&=\tan85^\circ\\ \tan55^\circ\cdot\tan65^\circ\cdot\cot15^\circ&=\tan85^\circ\\ \tan55^\circ\cdot\tan65^\circ\cdot\tan(90^\circ-15^\circ)&=\tan85^\circ\\ \color{blue}{\tan55^\circ\cdot\tan65^\circ\cdot\tan75^\circ}&\color{blue}{=\tan85^\circ}. \end{align} $$


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