Question

If $sin2A=\lambda ,sin2B$, then write the value of $\frac{\lambda +1}{\lambda -1}$

Answer 1

We have,
$sin2A=\lambda sin2B,\Rightarrow \frac{sin2A}{sin2B}=\lambda \Rightarrow \frac{sin2A}{sin2B}+1=\lambda +1\Rightarrow \frac{sin2A+sin2B}{sin2B}=\lambda +1\dots \left(i\right)Againsin2A=\lambda sin2B,\Rightarrow \frac{sin2A}{sin2B}=\lambda \Rightarrow \frac{sin2A}{sin2B}-1=\lambda -1\Rightarrow \frac{sin2A-sin2B}{sin2B}=\lambda -1\dots \left(ii\right)Dividingequation\left(i\right)byequation\left(ii\right),weget\frac{sin2A-sin2B}{sin2A-sin2B}=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{2sin\left(\frac{2A+2B}{2}\right)cos\left(\frac{2A-2B}{2}\right)}{2sin\left(\frac{2A-2B}{2}\right)cos\left(\frac{2A+2B}{2}\right)}=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{sin(A+B)cos(A-B)}{sin(A-B)cos(A+B)}=\frac{\lambda +1}{\lambda -1}\Rightarrow fracsin(A+B)cos(A-B)cos(A+B)sin(A-B)=\frac{\lambda +1}{\lambda -1}\Rightarrow \frac{tan(A+B)}{tan(A-B)}\frac{\lambda +1}{\lambda -1}\therefore \frac{\lambda +1}{\lambda -1}=\frac{tan(A+B)}{tan(A-B)}$