# Prove That: Sin ( a − B ) Cos a Cos B + Sin ( B − C ) Cos B Cos C + Sin ( C − a ) Cos C Cos a = 0 - Mathematics

Prove that:
$\frac{\sin \left( A - B \right)}{\cos A \cos B} + \frac{\sin \left( B - C \right)}{\cos B \cos C} + \frac{\sin \left( C - A \right)}{\cos C \cos A} = 0$

#### Solution

$\text{ LHS }= \frac{\sin\left( A - B \right)}{\cos A \cos B} + \frac{\sin\left( B - C \right)}{\cos B \cos C} + \frac{\sin\left( C - A \right)}{\cos C \cos A}$
$= \frac{\sin A \cos B - \cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C - \cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A - \cos C \sin A}{\cos C \cos A}$
$= \frac{\sin A \cos B}{\cos A \cos B} - \frac{\cos A \sin B}{\cos A \cos B} + \frac{\sin B \cos C}{\cos B \cos C} - \frac{\cos B \sin C}{\cos B \cos C} + \frac{\sin C \cos A}{\cos C \cos A} - \frac{\cos C \sin A}{\cos C \cos A}$
$= \frac{\sin A}{\cos A} - \frac{\sin B}{\cos B} + \frac{\sin B}{\cos B} - \frac{\sin C}{\cos C} + \frac{\sin C}{\cos C} - \frac{\sin A}{\cos A}$
$= \tan A - \tan B + \tan B - \tan C + \tan C - \tan A$
$= 0$
= RHS
Hence proved.

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 16.2 | Page 20