Prove that
sin (A + B + C) cos A sin B sin C
+ cos (A + B + C) sin A sin B sin C
- sin (A + B + C) cos A cos B cos C
- cos (A + B + C) sin A cos B cos C
+ sin (A + B) cos (B + C) cos (C + A)
+ cos (A + B) cos (B + C) sin (C + A) = 0

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Solution

Combine (1,3), (2,4), and (5,6),
L.H.S. $= - s i n (A + B + C) c o s A c o s (B + C) - c o s (A + B + C) s i n A c o s (b + C) + c o s (B + C) s i n (A + B + C + A) .$
Take out $c o s (B + C)$
L.H.S. = $c o s (B + C) s i n (A + B + C + A) - s i n (A + B + C + A) = 0$

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Q. Prove

sin (A + B + C) = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C - sin A sin B sin C

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

Q. Prove that:
(i)

\frac{s i n (A + B) + s i n (A - B)}{c o s (A + B) + c o s (A - B)} = t a n A

(ii)

\frac{s i n (A - B)}{c o s A c o s B} + \frac{s i n (B - C)}{c o s B c o s C} + \frac{s i n (C - A)}{c o s C c o s A} = 0

(iii)

\frac{s i n (A - B)}{s i n A s i n B} + \frac{s i n (B - C)}{s i n B s i n C} + \frac{s i n (C - A)}{s i n C s i n A} = 0

(iv)

s i n^{2} B = s i n^{2} A + s i n^{2} (A - B) - 2 s i n A c o s B s i n (A - B)

(v)

c o s^{2} + c o s^{2} B - 2 c o s A c o s B c o s (A + B) = s i n^{2} (A + B)

(vi)

\frac{t a n (A + B)}{c o t (A - B)} = \frac{t a n^{2} A - t a n^{2} B}{1 - t a n^{2} A t a n^{2} B}

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) $s i n (B - C) c o s (A - D) + s i n (C - A) c o s (B - D) + s i n (A - B) c o s (C - D) = 0$

Q. Is

cos (A + B + C) = cos A cos B cos C

, then

8 sin (B + C) sin (C + A) sin (A + B) + sin 2 A sin 2 B sin 2 C = 0

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Prove thatsin (A + B + C) cos A sin B sin C+ cos (A + B + C) sin A sin B sin C- sin (A + B + C) cos A cos B cos C- cos (A + B + C) sin A cos B cos C+ sin (A + B) cos (B + C) cos (C + A)+ cos (A + B) cos (B + C) sin (C + A) = 0

Combine (1,3), (2,4), and (5,6),L.H.S.=−sin(A+B+C)cosAcos(B+C)−cos(A+B+C)sinAcos(b+C)+cos(B+C)sin(A+B+C+A). Take out cos(B+C) L.H.S. = cos(B+C)sin(A+B+C+A)−sin(A+B+C+A)=0

Prove that
sin (A + B + C) cos A sin B sin C
+ cos (A + B + C) sin A sin B sin C
- sin (A + B + C) cos A cos B cos C
- cos (A + B + C) sin A cos B cos C
+ sin (A + B) cos (B + C) cos (C + A)
+ cos (A + B) cos (B + C) sin (C + A) = 0

Combine (1,3), (2,4), and (5,6),
L.H.S. $= - s i n (A + B + C) c o s A c o s (B + C) - c o s (A + B + C) s i n A c o s (b + C) + c o s (B + C) s i n (A + B + C + A) .$
Take out $c o s (B + C)$
L.H.S. = $c o s (B + C) s i n (A + B + C + A) - s i n (A + B + C + A) = 0$