Question

If $\alpha ={30}^{\circ }$ and $\beta ={60}^{\circ }$, then the value of $\frac{\sin \alpha +\sec 2\alpha +\tan (\alpha +{15}^{\circ })}{\tan \beta +\cot (\frac{\beta }{2}+{15}^{\circ })+\tan \alpha }$ is

• A. $3(4-\sqrt{3})$
• B. $\frac{7}{26}(4\sqrt{3}-3)$
• C. $\sqrt{3}-2$
• D. $\frac{1}{2}(\sqrt{3}-1)$

Answer 1

The correct option is B $\frac{7}{26}(4\sqrt{3}-3)$

$\frac{\sin \alpha +\sec 2\alpha +\tan (\alpha +{15}^{\circ })}{\tan \beta +\cot (\frac{\beta }{2}+{15}^{\circ })+\tan \alpha }$

$=\frac{\sin {30}^{\circ }+\sec {60}^{\circ }+\tan {45}^{\circ }}{\tan {60}^{\circ }+\cot {45}^{\circ }+\tan {30}^{\circ }}$

$=\frac{\frac{1}{2}+2+1}{\sqrt{3}+1+\frac{1}{\sqrt{3}}}$

$=\frac{7\sqrt{3}}{2(4+\sqrt{3})}=\frac{7}{26}(4\sqrt{3}-3)$