Question

Establish the identity

$\frac{\sin (\alpha +\beta )}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$

Short answer

By applying the sum formula of sin function we get $\frac{\sin (\alpha +\beta )}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$

Step-by-step solution

Step 1. Given information   

$\frac{\sin (\alpha +\beta )}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$

Step 2. Calculations  

$\frac{\sin (\alpha +\beta )}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$

take LHS, $\frac{\sin (\alpha +\beta )}{\tan \alpha +\tan \beta }$

$$\begin{gathered} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ \tan \alpha +\tan \beta =\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \beta }{\cos \beta }=\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta } \\ \frac{\sin (\alpha +\beta )}{\tan \alpha +\tan \beta }=\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta }} \\ =\cos \alpha \cos \beta \end{gathered}$$

LHS=RHS

Hence identity is established.