Question

If $A=B={60}^{\circ }$, verify that
(i) $\cos (A-B)=\cos A\cos B+\sin A\sin B$
(ii) $\sin (A-B)=\sin A\cos B-\cos A\sin B$
(iii) $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

Answer 1

$A=B={60}^{\circ }$

we know that

$\sin {60}^{\circ }=\frac{\sqrt{3}}{2},\sin {30}^{\circ }=\frac{1}{2}$

(i) $\cos (A-B)=\cos A\cos B+\sin A\sin B$

$\cos (60-60)=\cos 60\cos 60+\sin 60\sin 60$

$\cos \left(0\right)={\cos }^{2}60+{\sin }^{2}60$

$1={\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2$

$1=\left(\frac{1}{4}\right)+\left(\frac{3}{4}\right)$

$1=\frac{4}{4}$

$1=1$

Hence verified

(ii) $\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\sin (60-60)=\sin 60\cos 60-\cos 60\sin 60$

$\sin \left(0\right)=(\frac{\sqrt{3}}{2}\times \frac{1}{2})-(\frac{1}{2}\times \frac{\sqrt{3}}{2})$

$0=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$

$0=0$

Hence verified

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

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

$\tan \left(0\right)=\frac{0}{1+\tan 60\tan 60}$

$0=0$

Hence verified