Answer with Step-by-step explanation:
We are given that
[tex]tan(\alpha+\beta)[/tex]
[tex]\frac{sin(\alpha+\beta}{cos(\alpha+\beta)}[/tex]
By using formula[tex]tan x=\frac{sin x}{cos x}[/tex]
[tex]\frac{sin\alpha cos\beta+sin\beta cos\alpha}{cos\alpha cos\beta-sin\alpha sin\beta}[/tex]
By using property:[tex] sin(x+y)=sin x cosy+cos x sin y[/tex]
[tex] cos(x+y)=cos x cosy-sin x siny[/tex]
Divide numerator and denominator by [tex] cos\alpha cos\beta[/tex]
Then, we get
[tex]\frac{\frac{sin\alpha}{cos\alpha}+\frac{sin\beta}{cos\beta}}{1-\frac{sin\alpha sin\beta}{cos\alpha cos\beta}}[/tex]
[tex]tan(\alpha+\beta)=\frac{tan\alpha+tan\beta}{1-tan\alpha tan\beta}[/tex]
Hence, proved
Substitute [tex]\alpha=\beta[/tex]
Then we get
[tex]tan 2\alpha=\frac{tan\alpha+tan\alpha}{1-tan^2\alpha}[/tex]
[tex]tan(2\alpha)=\frac{2tan\alpha}{1-tan^2\alpha}[/tex]
Hence, proved.
