Question

derive the identity for \\( \tan ( \alpha + \beta ) \\) using \\( \tan ( \alpha + \beta ) = \frac { \sin ( \alpha + \beta ) } { \cos ( \alpha + \beta ) } \\). after applying the formulas for sums of sines and cosines, divide the numerator and denominator by \\( \cos \alpha \cos \beta \\).

divide the numerator and denominator by \\( \cos \alpha \cos \beta \\).

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

Explanation:

Step1: Apply sum - of - sines and sum - of - cosines formulas

We know that \(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\) and \(\cos(\alpha + \beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\). So, \(\tan(\alpha+\beta)=\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}\)

Step2: Divide numerator and denominator by \(\cos\alpha\cos\beta\)

For the numerator: \(\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta}=\frac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta}+\frac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}=\tan\alpha+\tan\beta\)
For the denominator: \(\frac{\cos\alpha\cos\beta-\sin\alpha\sin\beta}{\cos\alpha\cos\beta}=\frac{\cos\alpha\cos\beta}{\cos\alpha\cos\beta}-\frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}=1 - \tan\alpha\tan\beta\)

Answer:

\(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\)