# In Triangle Abc, Prove the Following: a − B a + B = Tan ( a − B 2 ) Tan ( a + B 2 ) - Mathematics

In triangle ABC, prove the following:

$\frac{a - b}{a + b} = \frac{\tan \left( \frac{A - B}{2} \right)}{\tan \left( \frac{A + B}{2} \right)}$

#### Solution

$\text{ Assume }\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$

Consider the LHS of the equation

$\frac{a - b}{a + b} = \frac{\tan \left( \frac{A - B}{2} \right)}{\tan \left( \frac{A + B}{2} \right)}$

$LHS = \frac{a - b}{a + b}$
$= \frac{k\left( \sin A - \sin B \right)}{k\left( \sin A + \sin B \right)}$

$\because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right), \sin A + \sin B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)$

$\therefore LHS = \frac{2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right)}{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}$
$= \frac{\tan\left( \frac{A - B}{2} \right)}{\tan\left( \frac{A + B}{2} \right)} = RHS$
$\text{ Hence proved } .$

Concept: Sine and Cosine Formulae and Their Applications
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 4 | Page 12