Advertisement Remove all ads

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: 

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

 

Advertisement Remove all ads

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?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 4 | Page 12
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×