Answer :
Answer:
(A)Quotient Identity
Step-by-step explanation:
To Prove: [tex]\dfrac{\sin \left(\alpha +\beta \right)}{\cos \left(\alpha +\beta \right)}=\tan \left(\alpha +\beta \right)[/tex]
Rewrite the numerator on the left side of the identity using the sum and difference formulas.
[tex]\dfrac{\sin \left(\alpha +\beta \right)}{\cos \left(\alpha +\beta \right)}=\dfrac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta -\sin \alpha \sin \beta }[/tex]
Next, Divide the numerator and denominator by[tex]cos\alpha \text{cos}\beta[/tex]
[tex]=\dfrac{\frac{\sin \alpha \cos \beta +\cos \alpha \sin \beta }{\cos \alpha \cos \beta }}{\frac{\cos \alpha \cos \beta -\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}[/tex]
Rewrite the fraction from the previous step such that it is a sum or difference of two expressions as shown below
[tex]=\dfrac{\frac{\sin \alpha{\cos \beta }}{\cos \alpha{\cos \beta }}+\frac{{\cos \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 }}[/tex]
Divide out any common factors in the expression from the previous step.
[tex]=\dfrac{\frac{\sin \alpha }{\cos \alpha }+\frac{\sin \beta }{\cos \beta }}{1-\frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}[/tex]
The expression from the previous step then simplifies to [tex]\tan \left(\alpha +\beta \right)[/tex] using the Quotient Identity
[tex]=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=\tan \left(\alpha +\beta \right)[/tex]
