Chapter 13.7, Problem 27WE

To determine

To Prove: The given identity.

Explanation of Solution

Given:

The given identity is $\cos \theta \cos \varphi \left(\tan \theta +\tan \varphi \right)=\sin \left(\theta +\varphi \right)$ .

Calculation:

Consider the given identity is,

  $\cos \theta \cos \varphi \left(\tan \theta +\tan \varphi \right)=\sin \left(\theta +\varphi \right)$

Consider the trigonometric identities.

  $\begin{array}{l}\sin \left(a+b\right)=\sin a\cos b+\cos a\sin b\\ \sin \left(a-b\right)=\sin a\cos b-\cos a\sin b\\ \cos \left(a+b\right)=\cos a\cos b-\sin a\sin b\\ \cos \left(a-b\right)=\cos a\cos b+\sin a\sin b\end{array}$

From, the given identity,

  $\begin{array}{c}\cos \theta \cos \varphi \left(\tan \theta +\tan \varphi \right)=\cos \theta \cos \varphi \left(\frac{\sin \theta }{\cos \theta }+\frac{\sin \varphi }{\cos \varphi }\right)\\ =\cos \theta \cos \varphi \left(\frac{\sin \theta \cos \varphi +\cos \theta \sin \varphi }{\cos \theta \cos \varphi }\right)\\ =\sin \theta \cos \varphi +\cos \theta \sin \varphi \\ =\sin \left(\theta +\varphi \right)\end{array}$

Hence, the given identity is proved.