diketahui tan A=⅓ tan B=¹/5 dengan A dan B sudut lancip hitunglah tan(2A+B) jawab

tan (2A+B)
[tex] =\frac{tan 2 \alpha + tan \beta \beta }{1-tan 2 \alpha (tan \beta)} [/tex]
[tex] =\frac{ \frac{2tan \alpha}{1- tan^{2} \alpha} + tan \beta }{1-\frac{2tan \alpha}{1- tan^{2} \alpha} (tan \beta)} [/tex]
[tex]= \frac{ \frac{2( \frac{1}{3})}{1- ( \frac{1}{3})^{2}} + \frac{1}{5}}{1-\frac{2( \frac{1}{3})}{1- ( \frac{1}{3})^{2}}(\frac{1}{5})} [/tex]
[tex]= \frac{ \frac{ \frac{2}{3}}{1- \frac{1}{9}} + \frac{1}{5}}{1-\frac{ \frac{2}{3}}{1- \frac{1}{9}} (\frac{1}{5})} [/tex]
[tex]=\frac{ \frac{ \frac{2}{3}}{\frac{8}{9}} + \frac{1}{5}}{1-\frac{ \frac{2}{3}}{\frac{8}{9}} (\frac{1}{5})}[/tex]
[tex]=\frac{ \frac{18}{24} + \frac{1}{5}}{1-\frac{18}{24} (\frac{1}{5})}[/tex]
[tex]= \frac{ \frac{90+24}{120}}{1-\frac{18}{120}}[/tex]
[tex]=(\frac{114}{120})( \frac{120}{102}})[/tex]
[tex]= \frac{19}{17}[/tex]

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