Prove that
$\frac{cos (A + B + C) + cos (- A + B + C) + cos (A - B + C) + cos (A + B - C)}{sin (A + B + C) + sin (- A + B + C) - sin (A - B + C) + sin (A + B - C)} = cot B$

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Solution

$L H S = \frac{(cos (A + B + C) + cos (- A + B + C)) + (cos (A - B + C) + cos (A + B - C))}{(sin (A + B + C) + sin (- A + B + C)) - (sin (A - B + C) - sin (A + B - C))}$
$= \frac{2 cos A cos (B + C) + 2 cos A cos (B - C)}{2 sin (B + C) cos A + 2 cos A sin (B - C)} = \frac{cos (B + C) + cos (B - C)}{sin (B + C) + sin (B - C)} = \frac{2 cos B cos C}{2 sin B cos C} = cot B = R H S$

Hence Proved

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Similar questions

Q. Prove that:
(i)

\frac{\cos (A + B + C) + \cos (- A + B + C) + \cos (A - B + C) + \cos (A + B - C)}{\sin (A + B + C) + \sin (- A + B + C) + \sin (A - B + C) - \sin (A + B - C)} = \cot C

(ii) sin (B − C) cos (A − D) + sin (C − A) cos (B − D) + sin (A − B) cos (C − D) = 0

Prove that:
(i) $\frac{c o s (A + B + C) + c o s (- A + B + C) + c o s (A - B + C) + c o s (A + B - C)}{s i n c o s (A + B + C) + s i n c o s (- A + B + C) + s i n c o s (A - B + C) - s i n (c o s (A + B - C))}$
(ii) $s i n (B - C) c o s (A - D) + s i n (C - A) c o s (B - D) + s i n (A - B) c o s (C - D) = 0$