please: in AB^Δ C prove that: ((cos(A))/(sin(B)sin(C))) +((cos(B))/(sin(A)sin(C)))+((cos(C))/(sin(A)sin(B))) =2


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Question Number 108573 by mnjuly1970 last updated on 17/Aug/20

$please : in A \overset{Δ}{B C} prove that :$ $\frac{c o s (A)}{s i n (B) s i n (C)} + \frac{c o s (B)}{s i n (A) s i n (C)} + \frac{c o s (C)}{s i n (A) s i n (B)} = 2$
	Answered by veth last updated on 17/Aug/20

	Commented by mnjuly1970 last updated on 17/Aug/20

	$thanks a lot master . mercey . . .$
	Answered by ajfour last updated on 17/Aug/20

	$t o p r o v e Σ \frac{\sin A \cos A}{\sin A \sin B \sin C} = 2$ $l . h . s . = \frac{1}{2} (\frac{Σ \sin 2 A}{Π \sin A})$ $= \frac{1}{2} (\frac{2 \sin (A + B) \cos (A - B) + \sin 2 C}{Π \sin A})$ $= \frac{\cos (A - B) + \cos C}{\sin A \sin B}$ $= 2 [\frac{\cos (A - B) + \cos C}{\cos (A - B) - \cos (A + B)}]$ $= 2 [\frac{\cos (A - B) + \cos C}{\cos (A - B) + \cos C}] = 2 .$
	Commented by mnjuly1970 last updated on 17/Aug/20

	$god bless you master . thanks a lot$
	Answered by nimnim last updated on 17/Aug/20

	$A + B + C = π .$ $L H S \frac{c o s A}{s i n B s i n C} - \frac{c o s (B + C)}{s i n A s i n C} - \frac{c o s (A + B)}{s i n A s i n B}$ $= \frac{c o s A}{s i n B s i n C} - [\frac{c o s A c o s C - s i n A s i n C}{s i n A s i n C} + \frac{c o s A c o s B - s i n A s i n B}{s i n A s i n B}]$ $= \frac{c o s A}{s i n B s i n C} - [\frac{c o s A c o s C}{s i n A s i n C} - 1 + \frac{c o s A c o s B}{s i n A s i n B} - 1]$ $= \frac{c o s A}{s i n B s i n C} - c o s A [\frac{s i n B c o s C + c o s B s i n C}{s i n A s i n B s i n C}] + 2$ $= \frac{c o s A}{s i n B s i n C} - \frac{c o s A s i n (B + C)}{s i n (B + C) s i n B s i n C} + 2$ $= \frac{c o s A}{s i n B s i n C} - \frac{c o s A}{s i n B s i n C} + 2$ $= 2 (R H S)$
	Commented by mnjuly1970 last updated on 17/Aug/20

	$thank you very much my dear$ $friend . peace be upon you . . .$
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