Question

Please prove this...​

Answer 1

Answer: see proof below

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 (k-1)/(k+1)\tan(\theta+\beta)/(2)`

`\text{Given:}\qquad \qquad ((\sin \theta)/(\sin \beta)-1)/((\sin \theta)/(\sin \beta)+1)}\cdot \tan(\theta +\beta)/(2)`

`\text{Simplify:}\qquad \qquad ((\sin \theta)/(\sin \beta)-(\sin \beta)/(\sin \beta))/((\sin \theta)/(\sin \beta)+(\sin \beta)/(\sin \beta))}\cdot \tan(\theta +\beta)/(2)\\\\\\.\qquad \qquad \qquad = (\sin \theta -\sin \beta)/(\sin \theta +\sin \beta)\cdot \tan(\theta +\beta)/(2)`

`\text{Product to Sum:}\qquad \quad (2\cos (\theta+\beta)/(2)\cdot \sin (\theta-\beta)/(2))/(2\sin (\theta+\beta)/(2)\cdot \cos (\theta-\beta)/(2))\cdot \tan(\theta +\beta)/(2)`

`\text{Expand:}\qquad \qquad (2\cos (\theta+\beta)/(2)\cdot \sin (\theta-\beta)/(2))/(2\sin (\theta+\beta)/(2)\cdot \cos (\theta-\beta)/(2))\cdot (\sin(\theta +\beta)/(2))/(\cos (\theta +\beta)/(2))`

`\text{Simplify:}\qquad \qquad \quad \quad (\sin (\theta +\beta)/(2))/(\cos (\theta+ \beta)/(2))\\\\\\.\qquad \qquad \qquad \quad =\tan(\theta +\beta)/(2)`

`\text{LHS = RHS:}\quad \tan(\theta +\beta)/(2)=\tan(\theta +\beta)/(2)\quad \checkmark`