Question

Evaluate each of the following :
$2si{n}^2{30}^{\circ }-3co{s}^2{45}^{\circ }+ta{n}^2{60}^{\circ }$

Answer 1

Given $2{\sin }^2{30}^{\circ }-3{\cos }^2{45}^{\circ }+{\tan }^2{60}^{\circ }$

$=2{\left(\frac{1}{2}\right)}^2-3{\left(\frac{1}{\sqrt{2}}\right)}^2+{\left(\sqrt{3}\right)}^2$ $[\because \cos {45}^{\circ }=\frac{1}{\sqrt{2}},\sin {30}^{\circ }=\frac{1}{2},\tan {60}^{\circ }=\sqrt{3}]$

$=\frac{2}{4}-\frac{3}{2}+3$

$=\frac{1}{2}-\frac{3}{2}+\frac{3\times 2}{2}$

$=\frac{1}{2}-\frac{3}{2}+\frac{6}{2}$

$=\frac{4}{2}$

$=2$

Hence, $2{\sin }^2{30}^{\circ }-3{\cos }^2{45}^{\circ }+{\tan }^2{60}^{\circ }=2$