Question

$\cos {60}^{\circ }\cos {30}^{\circ }-\sin {60}^{\circ }\sin {30}^{\circ }$ is equal to

• A. $\cos {90}^{\circ }$
• B. $cosec{90}^{\circ }$
• C. $\sin {30}^{\circ }+\cos {30}^{\circ }$
• D. $\tan {90}^{\circ }$

Answer 1

The correct option is A $\cos {90}^{\circ }$

$\cos 60°\cos 30°-\sin 60°\sin 30°=\cos 90°$

$=\frac{1}{2}\times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\times \frac{1}{2}$

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

$=0$

Also, $\cos 90°=1$

$\therefore \cos 60°\cos 30°-\sin 60°\sin 30°=\cos 90°$

Hence, the answer is $\cos 90°.$