Question

$\frac{\tan {60}^o-\tan {30}^o}{1+\tan {60}^o\tan {30}^o}+\cos {60}^o\cos {30}^o+\sin {60}^o\sin {30}^o$

Answer 1

Using compound angles formula, wehave

$\frac{\tan {60}^{\circ }-\tan {30}^{\circ }}{1+\tan {60}^{\circ }\tan {30}^{\circ }}+\cos {60}^{\circ }\cos {30}^{\circ }+\sin {60}^{\circ }\sin {30}^{\circ }$

$=\tan ({60}^{\circ }-{30}^{\circ })+\cos ({60}^{\circ }-{30}^{\circ })$

$=\tan {30}^{\circ }+\cos {30}^{\circ }$

$=\sqrt{3}+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$