Question

$2{\tan }^2{45}^o+3{\cos }^2{30}^o-{\sin }^2{60}^o=\frac{p}{2}$, $p$ is

Answer 1

$2{\tan }^2(45{)}^{\circ }+3{\cos }^2{30}^{\circ }-{\sin }^2{60}^{\circ }$
$=2\left(1\right)+3\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)$
$\frac{8+9-3}{4}=\frac{7}{2}=\frac{p}{2}$
$\therefore p=7$