Chapter 7, Problem 14CT

To determine

Toprove: The identity $\frac{\sin \left(\alpha +\beta \right)}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$ .

Explanation of Solution

Given information:

Instruction is to establish the identity $\frac{\sin \left(\alpha +\beta \right)}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$

Formula used:

1) $\tan \theta =\frac{\sin \theta }{\cos \theta }$

2) $\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

Proof:

Consider L.H.S. of the identity,

  $\frac{\sin \left(\alpha +\beta \right)}{\tan \alpha +\tan \beta }$

By using $\tan \theta =\frac{\sin \theta }{\cos \theta }$ ,

  $\Rightarrow \frac{\sin \left(\alpha +\beta \right)}{\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \beta }{\cos \beta }}$

By using cross multiplication,

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

By using $\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta$

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

By simplifying this,

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

Cancelling $\sin \left(\alpha +\beta \right)$ from the numerator and denominator,

  $=\cos \alpha \cos \beta$

  = R.H.S.

Hence, the identity $\frac{\sin \left(\alpha +\beta \right)}{\tan \alpha +\tan \beta }=\cos \alpha \cos \beta$ is proved.