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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

$$\tan (\mathrm{A}+\mathrm{B})$$$$=\frac{\sin (\mathrm{A}+\mathrm{B})}{\cos (\mathrm{A}+\mathrm{B})}=\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}-\mathrm{sinASinB}}$$$$=\frac{\frac{\mathrm{sinAcosB}+\mathrm{cosAsinB}}{\mathrm{cosAcosB}}}{\frac{\mathrm{cosAcosB}-\mathrm{sinAsinB}}{\mathrm{cosAcosB}}}$$$$=\frac{\mathrm{tanA}+\mathrm{tanB}}{1-\mathrm{tanAtanB}}$$$$\mathrm{now}$$$$\frac{1+\tan 15}{1-\tan 15}$$$$=\frac{\tan 45+\tan 15}{1-\tan 45\tan 15}$$$$=\tan (45+15)=\tan 60=\sqrt{3}\mathrm{ans}$$

Commented by ZiYangLee last updated on 16/Nov/20

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

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