# If Sin 2a = λ Sin 2b, Then Write the Value of λ + 1 λ − 1 - Mathematics

Sum

If sin 2A = λ sin 2B, then write the value of $\frac{\lambda + 1}{\lambda - 1}$

#### Solution

Given:
sin 2A = λ sin 2B

$\frac{\lambda + 1}{\lambda - 1}$
$\Rightarrow \frac{\sin2A + \sin2B}{\sin2A - \sin2B} = \frac{\lambda + 1}{\lambda - 1}$
$\Rightarrow \frac{2\sin\left( \frac{2A + 2B}{2} \right)\cos\left( \frac{2A - 2B}{2} \right)}{2\sin\left( \frac{2A - 2B}{2} \right)\cos\left( \frac{2A + 2B}{2} \right)} = \frac{\lambda + 1}{\lambda - 1} \left[ \because \sin A + \sin B = 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \text{and }\sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \right]$
$\Rightarrow \frac{\sin\left( A + B \right)\cos\left( A - B \right)}{\sin\left( A - B \right)\cos\left( A + B \right)} = \frac{\lambda + 1}{\lambda - 1}$
$\Rightarrow \tan\left( A + B \right)\cot\left( A - B \right)=\frac{\lambda + 1}{\lambda - 1}$
$\Rightarrow\frac{\tan\left( A + B \right)}{\tan\left( A - B \right)}=\frac{\lambda + 1}{\lambda - 1}$

Concept: Transformation Formulae
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Q 8 | Page 21