If A = 60° and B = 30°, verify that:


If A = 60° and B = 30°, verify that:

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



(i) $\sin (A-B)=\sin 30^{\circ}=\frac{1}{2}$

sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o

$=\left(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}\right)=\left(\frac{3}{4}-\frac{1}{4}\right)=\frac{2}{4}=\frac{1}{2}$

∴ sin (A − B) = sin A cos B − cos A sin B

(ii) $\cos (A-B)=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$

cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o

$=\left(\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\right)=2 \times \frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$

∴​ cos (A − B) = cos A cos B + sin A sin B

(iii) $\tan (A-B)=\tan 30^{\circ}=\frac{1}{\sqrt{3}}$

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

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





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