Question

please prove this...​

Answer 1

Answer:  see proof below

Step-by-step explanation:

Use the following Sum to Product Identities:

sin (A) - sin (B) = 2 cos (A+B)/2 · sin (A-B)/2

sin (A) + sin (B) = 2 sin (A+B)/2 · cos (A-B)/2

Given:  sin Ф = k sin β      -->   (sin Ф)/(sin β) = k

Proof RHS → LHS

$\text{RHS:}\qquad \qquad \qquad \dfrac{k-1}{k+1}\tan\dfrac{\theta+\beta}{2}$

$\text{Given:}\qquad \qquad \dfrac{\frac{\sin \theta}{\sin \beta}-1}{\frac{\sin \theta}{\sin \beta}+1}}\cdot \tan\dfrac{\theta +\beta}{2}$

$\text{Simplify:}\qquad \qquad \dfrac{\frac{\sin \theta}{\sin \beta}-\frac{\sin \beta}{\sin \beta}}{\frac{\sin \theta}{\sin \beta}+\frac{\sin \beta}{\sin \beta}}}\cdot \tan\dfrac{\theta +\beta}{2}\\\\\\.\qquad \qquad \qquad = \dfrac{\sin \theta -\sin \beta}{\sin \theta +\sin \beta}\cdot \tan\dfrac{\theta +\beta}{2}$

$\text{Product to Sum:}\qquad \quad \dfrac{2\cos \frac{\theta+\beta}{2}\cdot \sin \frac{\theta-\beta}{2}}{2\sin \frac{\theta+\beta}{2}\cdot \cos \frac{\theta-\beta}{2}}\cdot \tan\dfrac{\theta +\beta}{2}$

$\text{Expand:}\qquad \qquad \dfrac{2\cos \frac{\theta+\beta}{2}\cdot \sin \frac{\theta-\beta}{2}}{2\sin \frac{\theta+\beta}{2}\cdot \cos \frac{\theta-\beta}{2}}\cdot \dfrac{\sin\frac{\theta +\beta}{2}}{\cos \frac{\theta +\beta}{2}}$

$\text{Simplify:}\qquad \qquad \quad \quad \dfrac{\sin \frac{\theta +\beta}{2}}{\cos \frac{\theta+ \beta}{2}}\\\\\\.\qquad \qquad \qquad \quad =\tan\dfrac{\theta +\beta}{2}$

 $\text{LHS = RHS:}\quad \tan\dfrac{\theta +\beta}{2}=\tan\dfrac{\theta +\beta}{2}\quad \checkmark$