Question

Prove that ${\tan }^2{30}^{\circ }+2\sin {60}^{\circ }+\tan {45}^{\circ }-\tan {60}^{\circ }+{\cos }^2{30}^{\circ }=\frac{25}{12}$

Answer 1

$\Rightarrow {\tan }^2{30}^0+2\sin {60}^0+\tan {45}^0-\tan {60}^0+{\cos }^2{30}^0$

$\Rightarrow \frac{1}{3}+2\left(\frac{\sqrt{3}}{2}\right)+1-\sqrt{3}+\frac{3}{4}$

$\Rightarrow \frac{25}{12}$