Respuesta :
Answer:
Proved
Step-by-step explanation:
To Prove: [tex]tan(\alpha+\beta) =\dfrac{tan(\alpha)+tan(\beta)}{1 + tan(\alpha)tan(\beta)}[/tex]
Proof:
Now: [tex]tan \theta =\dfrac{sin\theta }{cos \theta}[/tex]
Therefore:
[tex]tan (\alpha+\beta)=\dfrac{ sin (\alpha+\beta)}{cos (\alpha+\beta) }[/tex]
Applying these angle sum formula
[tex]sin (\alpha+\beta)=sin \alpha cos \alpha + sin \beta cos \beta\\cos (\alpha+\beta)=cos \alpha cos \beta - sin \alpha sin \beta[/tex]
[tex]tan (\alpha+\beta)=\dfrac{ sin \alpha cos \alpha + sin \beta cos \beta}{cos \alpha cos \beta - sin \alpha sin \beta }[/tex]
Divide all through by [tex]cos \alpha cos \beta[/tex]
[tex]tan (\alpha+\beta)=\dfrac{ (sin \alpha cos \alpha)/(cos \alpha cos \beta) + (sin \beta cos \beta)/(cos \alpha cos \beta)}{(cos \alpha cos \beta)/(cos \alpha cos \beta) - (sin \alpha sin \beta)/(cos \alpha cos \beta) }\\\\tan (\alpha+\beta)=\dfrac{\frac{sin \alpha}{cos \alpha}+\frac{sin \beta}{cos \beta} }{1-tan \alpha tan \beta} \\$Therefore:\\\\tan (\alpha+\beta)=\dfrac{tan \alpha+tan \beta}{1-tan \alpha tan \beta}[/tex]
=sin \alpha cos \beta + cos \alpha sin \beta/cos \alpha cos \beta/cos \alpha cos \beta- sin \alpha sin \beta/cos \alpha cos \beta
=sin \alpha/cos \alpha + sin \beta/cos \beta/1-tan \alpha tan \beta
=tan A + tan B/1-tan A tan B
