Question

Find the values of $\frac{\tan {60}^o-\tan {30}^o}{1+\tan {60}^o.\tan {30}^o}+(\sin {30}^o.\cos {60}^o+\cos {30}^o.\sin {60}^o)$.

Answer 1

Now,

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

$=\tan ({60}^o-{30}^o)+\sin ({30}^o+{60}^o)$ [ Using the direct formulas for $\tan (A-B)$ and $\sin (A+B)$]

$=\tan {30}^o+\sin {90}^o$

$=\frac{1}{\sqrt{3}}+1$

$=\frac{\sqrt{3}+1}{\sqrt{3}}$.