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In a triangle ABC, if $\cot(A)+\cot(B)+\cot(C)=0$ then find value of $$\cos(A)\cos(B)\cos(C)$$

i tried to take LCMs but couldnot reach near to target

Thanks for helping me out

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    See also : http://math.stackexchange.com/questions/1849467/given-abc-180-circ-find-value-of-tan-a-cdot-tan-b-tan-b-cdot-tan-c-t?rq=1 – 2017-02-07

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$$\cot^2A+\cot^2B+\cot^2C$$

$$=(\cot A+\cot B+\cot C)^2-2(\cot A\cot B+\cot B\cot C+\cot C\cot A)$$

Now use Proving $\cot(A)\cot(B)+\cot(B)\cot(C)+\cot(C)\cot(A)=1$ to disprove the proposed condition.

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    Can you add more details.thanks – 2017-02-07
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    How can sum of sqaures equals to -2 – 2017-02-07
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    @sophie exactly, it can't .. so the condition given never holds. – 2017-02-07
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    But how to find value of product – 2017-02-07
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    @SophieClad, How can we find the product if the given condition itself is impossible. Btw, what is the source of the problem? – 2017-02-07
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    @labbhattacharjee you mean sum of cot's can never be 0.i wonder textbook states answer to be -1 – 2017-02-07
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    @SophieClad, For real triangle, $|\cos x|<1$ for $x=A, B,C$ So, $$|\cos A\cos B\cos C|<1\implies\cos A\cos B\cos C>-1$$ – 2017-02-07
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    is your last inequality true? shouldn't there be < instead? – 2017-02-07
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    @SophieClad, It is rightly greater than $-1$. As $|a|<1\implies$ $$-1 – 2017-02-07
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    @labbhattacharjee Can we say that product of any number of sin or cos or both can never exceed 1 and always greater than -1 – 2017-02-07
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53167/discussion-between-lab-bhattacharjee-and-sophie-clad). – 2017-02-07
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    @labbhattacharjee okay kindly continue. – 2017-02-07