Find the value of $$S=\cot(16)\cot(44)+\cot(44)\cot(76)-\cot(76)\cot(16)$$

Note:All angles are in degrees

My method:

I used the identity


So choosing $x=16$ we get

$$\tan76 \tan44 \tan 16=\tan48$$







can i need a clue from here?


The idea is to exploit the identity $$\cot(\alpha+\beta) = \frac{\cot \alpha \cot \beta - 1}{\cot \alpha + \cot\beta}.$$ With $\alpha = 16^\circ$, $\beta = 44^\circ$, we get for instance $$\cot 16^\circ \cot 44^\circ = (\cot 16^\circ + \cot 44^\circ)\cot 60^\circ + 1.$$ Similarly, $$\cot 44^\circ \cot 76^\circ = (\cot 44^\circ + \cot 76^\circ) \cot 120^\circ + 1,$$ and $$\cot (-16^\circ) \cot 76^\circ = (\cot (-16^\circ) + \cot 76^\circ) \cot 60^\circ + 1.$$ Note $\cot (-\theta) = -\cot \theta$. All that is left to compute the sum of these three expressions, observing that $\cot 60^\circ = - \cot 120^\circ$.


Let us see how the problem came into being in my understanding.

From Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$,

if $\cot3x=\cot3y$ for non-zero finite $\tan x,$

$$\tan^3x-3\tan3y\tan^2x-3\tan x+\tan3y=0$$

$$\iff\cot^3x\tan3y-3\cot^2x-3\tan3y\cot x+1=0$$

whose roots are $\cot\left(x+60^\circ n\right), n=0,1,2$

$$\implies\cot x\cot\left(x+60^\circ\right)+\cot\left(x+60^\circ\right)\cot\left(x+120^\circ\right)+\cot\left(x+120^\circ\right)\cot x=\dfrac{-3\tan3y}{\tan3y}$$

$=-3$ for for non-zero finite $\tan3y$

Set $x=16^\circ$ and $\cot\left(16^\circ+120^\circ\right)=\cot(180^\circ-44^\circ)=-\cot44^\circ$


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