please-in-AB-C-prove-that-cos-A-sin-B-sin-C-cos-B-sin-A-sin-C-cos-C-sin-A-sin-B-2-

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 \dots

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 \dots