Question

Find the value of the following:
$2{\sin }^2{30}^o-3{\cos }^2{30}^o+\tan {60}^o+3{\sin }^2{90}^o$

Answer 1

We know that the values of the trignometric functions $\sin {30}^0=\frac{1}{2}$, $\cos {30}^0=\frac{\sqrt{3}}{2}$, $\tan {60}^0=\sqrt{3}$ and $\sin {90}^0=1$, therefore, the value of $2{\sin }^2{30}^0-3{\cos }^2{30}^0+\tan {60}^0+3{\sin }^2{90}^0$ is:

$2{\sin }^2{30}^0-3{\cos }^2{30}^0+\tan {60}^0+3{\sin }^2{90}^0=2\times {\left(\frac{1}{2}\right)}^2-3\times {\left(\frac{\sqrt{3}}{2}\right)}^2+\sqrt{3}+3\times 1^2$

$=(2\times \frac{1}{4})-(3\times \frac{3}{4})+\sqrt{3}+3=\frac{1}{2}-\frac{9}{4}+3+\sqrt{3}=\frac{(2-9+12)+4\sqrt{3}}{4}=\frac{5+4\sqrt{3}}{4}$

Hence, $2{\sin }^2{30}^0-3{\cos }^2{30}^0+\tan {60}^0+3{\sin }^2{90}^0=\frac{5+4\sqrt{3}}{4}$.