Question

tan A+ tan (60+A) -tan (60-A) is equal to

• A. 3 tan 3A
• B. tan 3A
• C. cot 3A
• D. sin 3A

Answer 1

The correct option is A 3 tan 3A

$\tan (60+A)=\frac{\tan {60}^o+\tan A}{1-\tan {60}^o\tan A}=\frac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}$

$\tan (60-A)=\frac{\tan {60}^o-\tan A}{1+\tan {60}^o\tan A}=\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}$

$\tan (60+A)-\tan (60-A)=\frac{\sqrt{3}+\tan A}{1-\sqrt{3}\tan A}-\frac{\sqrt{3}-\tan A}{1+\sqrt{3}\tan A}$

$=\frac{(\sqrt{3}+\tan A)(1+\sqrt{3}\tan A)-(\sqrt{3}-\tan A)(1-\sqrt{3}\tan A)}{1^2-(\sqrt{3}\tan A{)}^2}$

$=\frac{\sqrt{3}+3\tan A+\tan A+\sqrt{3}{\tan }^{2}A-\sqrt{3}+3\tan A+\tan A-\sqrt{3}{\tan }^{2}A}{1-3{\tan }^{2}A}$

$=\frac{8\tan A}{1-3{\tan }^{2}A}$

$\tan A+\tan (60+A)-\tan (60-A)=\tan A+\frac{8\tan A}{1-3{\tan }^{2}A}$

$=\frac{\tan A-3{\tan }^{3}A+8\tan A}{1-3{\tan }^{2}A}$

$=\frac{3(3\tan A-{\tan }^{3}A)}{1-3{\tan }^{2}A}$

$=3\tan 3A$

$(\because \tan 3A=\frac{3\tan A-{\tan }^{3}A}{1-3{\tan }^{2}A})$.