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Range of Trigonometric Ratios from 0 to 90 Degrees

Prove that Si...

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Prove that SinA+sinB=sin(A+B)cos(A+B)/2

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sin (A + B) \neq sin A + sin B

Q. Prove that:

(i)

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

\frac{\sin (A - B)}{\cos A \cos B} + \frac{\sin (B - C)}{\cos B \cos C} + \frac{\sin (C - A)}{\cos C \cos A} = 0

\frac{\sin (A - B)}{\sin A \sin B} + \frac{\sin (B - C)}{\sin B \sin C} + \frac{\sin (C - A)}{\sin C \sin A} = 0

(iv) sin² B = sin² A + sin² (A − B) − 2 sin A cos B sin (A − B)
(v) cos² A + cos² B − 2 cos A cos B cos (A + B) = sin² (A + B)
(vi)

\frac{\tan (A + B)}{\cot (A - B)} = \frac{\tan^{2} A - \tan^{2} B}{1 - \tan^{2} A \tan^{2} B}

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

A + B + C = π

(sin A + sin B + sin C) (- sin A + sin B + sin C) (sin A - sin B + sin C) (sin A + sin B - sin C) = 4 {sin}^{2} A {sin}^{2} B {sin}^{2} C

\frac{a^{2} sin (B - C)}{sin B + sin C} + \frac{b^{2} sin (C - A)}{sin C + sin A} + \frac{c^{2} sin (A - B)}{sin A + sin B} = 0

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