Prove that $ 1+ \cos A + \cos B + \cos C = 0$.

Asked Jul 02 '14 at 07:46

Active Jul 02 '14 at 07:49

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If $A+B+C=180^\circ$ and

$\tan \left[\dfrac{A+B-C} 4 \right] \tan \left[ \dfrac{-A+B+C} 4\right] \tan\left[\dfrac{A-B+C} 4 \right] =1$

then prove that

$ 1+ \cos A + \cos B + \cos C = 0$

I tried to use the formula of $\tan(A+B+C)$, but couldn't solve it.

edited Jul 02 '14 at 07:49

M. Vinay

asked Jul 02 '14 at 07:46

maths lover

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http://math.stackexchange.com/questions/624189/prove-that-1-cos-alpha-cos-beta-cos-gamma-0 – lab bhattacharjee Jul 02 '14 at 08:58

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