Question

If A = 60° and B = 30°, verify that:

(i) sin (A − B) = sin A cos B − cos A sin B
(ii) cos (A − B) = cos A cos B + sin A sin B
(iii) tan (A − B) = $\frac{\tan A-\tan B}{1+\tan A\tan B}$

Answer 1

(i) sin (A − B) = sin 30^o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60^o cos 30^o − cos 60^o sin 30^o
= $\left(\frac{\sqrt{3}}{2}\times \frac{\sqrt{3}}{2}-\frac{1}{2}\times \frac{1}{2}\right)=\left(\frac{3}{4}-\frac{1}{4}\right)=\frac{2}{4}=\frac{1}{2}$
∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (A − B) = cos 30^o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60^o cos 30^o + sin 60^o sin 30^o
= $\left(\frac{1}{2}\times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\times \frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\right)=2\times \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$
∴​ cos (A − B) = cos A cos B + sin A sin B

(iii) tan (A − B) = tan 30^o = $\frac{1}{\sqrt{3}}$
$\frac{\tan A-\tan B}{1+\tan A\tan B}=\frac{\tan {60}^o-\tan {30}^o}{1+\tan {60}^o\tan {30}^o}=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\left(\sqrt{3}\times \frac{1}{\sqrt{3}}\right)}=\frac{1}{2}\times \left(\frac{3-1}{\sqrt{3}}\right)=\frac{1}{\sqrt{3}}$
∴​ tan (A − B) = $\frac{\tan A-\tan B}{1+\tan A\tan B}$