SinA-B cos A cos B -sinB-C cos B cos C -sinC-A cos A cos C -

sin (A B)cos A cos B + sin (B C)cos B cos C + sin (C A)cos C cos A = sin A cos B cos A sin Bcos A cos B + sin B cos C cos B sin Ccos B cos C ...

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Trigonometric Ratios of Allied Angles

sinA-B cos A ...

Question

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

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A, B, C \neq \frac{(2 n + 1) π}{2}

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

\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

\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

\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

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)

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)

\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}

A, B, C \neq \frac{(2 n + 1) π}{2}

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

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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