Question

Show that: $\dfrac{{\sin (A - B)}}{{\cos A\cos B}} + \dfrac{{\sin (B - C)}}{{\cos B\cos C}} + \dfrac{{\sin (C - A)}}{{\cos C\cos A}} = 0$

Answer 1

Hint : 1. The expansion of sin(A-B) can be written as
$\sin (A - B) = \sin A\cos B - \cos A\sin B$
2. $\dfrac{{\sin A}}{{\cos A}} = \tan A$

Complete step-by-step answer :
1. We are provided with the expression
$\dfrac{{\sin (A - B)}}{{\cos A\cos B}} + \dfrac{{\sin (B - C)}}{{\cos B\cos C}} + \dfrac{{\sin (C - A)}}{{\cos C\cos A}}$
2. Applying the expansion of sin(A-B) in the above expression, we will get
$\dfrac{{\sin A\cos B - \cos A\sin B}}{{\cos A\cos B}} + \dfrac{{\sin B\cos C - \cos B\sin C}}{{\cos B\cos C}} + \dfrac{{\sin C\cos A - \cos C\sin A}}{{\cos C\cos A}}$
3. Since the denominator is common, now splitting the numerator and we will get
$\dfrac{{\sin A\cos B}}{{\cos A\cos B}} - \dfrac{{\cos A\sin B}}{{\cos A\cos B}} + \dfrac{{\sin B\cos C}}{{\cos B\cos C}} - \dfrac{{\cos B\sin C}}{{\cos B\cos C}} + \dfrac{{\sin C\cos A}}{{\cos C\cos A}} - \dfrac{{\cos C\sin A}}{{\cos C\cos A}}$
4. Eliminating the term which are same in numerator and denominator, and we will get
$\dfrac{{\sin A}}{{\cos A}} - \dfrac{{\sin B}}{{\cos B}} + \dfrac{{\sin B}}{{\cos B}} - \dfrac{{\sin C}}{{\cos C}} + \dfrac{{\sin C}}{{\cos C}} - \dfrac{{\sin A}}{{\cos A}}$
5 Now, since we know$\dfrac{{\sin A}}{{\cos A}} = \tan A$. Applying this in the above expression
i.e. $\tan A - \tan B + \tan B - \tan C + \tan C - \tan A = 0$
Hence, proved that
$\dfrac{{\sin (A - B)}}{{\cos A\cos B}} + \dfrac{{\sin (B - C)}}{{\cos B\cos C}} + \dfrac{{\sin (C - A)}}{{\cos C\cos A}} = 0$

Note : Point to be considered in the above question is that when we are provided a number in the fractional form, and the numerator can be splitted into two parts while keeping the denominator the same.