Prove that: (i) sin(A+B)+sin(A−B)cos(A+B)+cos(A−B)=tanA (ii) sin(A−B)cosAcosB+sin(B−C)cosBcosC+sin(C−A)cosCcosA=0 (iii) sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA=0 (iv) sin2B=sin2A+sin2(A−B)−2sinAcosBsin(A−B) (v) cos2+cos2B−2cosAcosBcos(A+B)=sin2(A+B) (vi) tan(A+B)cot(A−B)=tan2A−tan2B1−tan2Atan2B

Prove that: (i) sin(A+B)+sin(A−B)cos(A+B)+cos(A−B)=tanA (ii) sin(A

1.	Prove that: (i) sin(A+B)+sin(A−B)cos(A+B)+cos(A−B)=tanA (ii) sin(A−B)cosAcosB+sin(B−C)cosBcosC+sin(C−A)cosCcosA=0 (iii) sin(A−B)sinAsinB+sin(B−C)sinBsinC+sin(C−A)sinCsinA=0 (iv) sin2B=sin2A+sin2(A−B)−2sinAcosBsin(A−B) (v) cos2+cos2B−2cosAcosBcos(A+B)=sin2(A+B) (vi) tan(A+B)cot(A−B)=tan2A−tan2B1−tan2Atan2B
Answer» 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) $\frac{t a n (A + B)}{c o t (A - B)} = \frac{t a n^{2} A - t a n^{2} B}{1 - t a n^{2} A t a n^{2} B}$