Chapter 13.7, Problem 28WE

To determine

To Prove: The given identity.

Explanation of Solution

Given:

The given identity is $\sin \theta \sin \varphi \left(\cot \theta \cot \varphi -1\right)=\cot \left(\theta +\varphi \right)$ .

Calculation:

Consider the given identity is,

  $\sin \theta \sin \varphi \left(\cot \theta \cot \varphi -1\right)=\cot \left(\theta +\varphi \right)$

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,

  $\begin{array}{c}\sin \theta \sin \varphi \left(\cot \theta \cot \varphi -1\right)=\sin \theta \sin \varphi \left(\frac{\cos \theta }{\sin \theta }\frac{\cos \varphi }{\sin \varphi }-1\right)\\ =\sin \theta \sin \varphi \left(\frac{\cos \theta \cos \varphi -\sin \theta \sin \varphi }{\sin \theta \sin \varphi }\right)\\ =\cos \theta \cos \varphi -\sin \theta \sin \varphi \\ =\cos \left(\theta +\varphi \right)\end{array}$

Hence, the given identity is proved.