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Question Number 122295 by ZiYangLee last updated on 15/Nov/20

Answered by som(math1967) last updated on 15/Nov/20

tan(A+B)  =((sin(A+B))/(cos(A+B)))=((sinAcosB+cosAsinB)/(cosAcosB−sinASinB))  =(((sinAcosB+cosAsinB)/(cosAcosB))/((cosAcosB−sinAsinB)/(cosAcosB)))  =((tanA+tanB)/(1−tanAtanB))  now  ((1+tan15)/(1−tan15))  =((tan45+tan15)/(1−tan45tan15))  =tan(45+15)=tan60=(√3) ans

$$\mathrm{tan}\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{A}+\mathrm{B}\right)}{\mathrm{cos}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}−\mathrm{sinASinB}} \\ $$$$=\frac{\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}}}{\frac{\mathrm{cosAcosB}−\mathrm{sinAsinB}}{\mathrm{cosAcosB}}} \\ $$$$=\frac{\mathrm{tanA}+\mathrm{tanB}}{\mathrm{1}−\mathrm{tanAtanB}} \\ $$$$\mathrm{now} \\ $$$$\frac{\mathrm{1}+\mathrm{tan15}}{\mathrm{1}−\mathrm{tan15}} \\ $$$$=\frac{\mathrm{tan45}+\mathrm{tan15}}{\mathrm{1}−\mathrm{tan45tan15}} \\ $$$$=\mathrm{tan}\left(\mathrm{45}+\mathrm{15}\right)=\mathrm{tan60}=\sqrt{\mathrm{3}}\:\mathrm{ans} \\ $$

Commented by ZiYangLee last updated on 16/Nov/20

thanks

$$\mathrm{thanks} \\ $$$$ \\ $$

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