Question

If A = 60⁰ and B = 30⁰, verify that: 

(iii) tan (A-B) = ${\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}}$

Answer 1

tan (A – B) = tan 60^{0 =${\displaystyle \frac{1}{\sqrt{3}}}$}

${\displaystyle \frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{{\tan 60}^0-{\tan 30}^{60}}{1+{\tan 60}^0{\tan 30}^0}=\frac{\sqrt{3}-\left(\frac{1}{\sqrt{3}}\right)}{1+(\sqrt{3}\times \frac{1}{\sqrt{3}})}}$  = ${\displaystyle \frac{1}{2}\times \frac{3-1}{\sqrt{3}}=\frac{3}{\sqrt{3}}}$

∴ ${\displaystyle \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}}$