Question

$2ta{n}^2{45}^{\circ }+3co{s}^2{30}^{\circ }-si{n}^2{60}^{\circ }$ = ________ .

Answer 1

The value of $tan{45}^{\circ }=1,cos{30}^{\circ }=\frac{\sqrt{3}}{2}$
and $sin{60}^{\circ }=\frac{\sqrt{3}}{2}$
Then, by substituiting the values we get,
$\Rightarrow 2ta{n}^2{45}^{\circ }+3co{s}^2{30}^{\circ }-si{n}^2{60}^{\circ }=2(1{)}^2+3(\frac{\sqrt{3}}{2}{)}^2-(\frac{\sqrt{3}}{2}{)}^2=2+\frac{9}{4}-\frac{3}{4}=\frac{8+9-3}{4}=\frac{14}{4}=\frac{7}{2}$
So, $2ta{n}^2{45}^{\circ }+3co{s}^2{30}^{\circ }-si{n}^2{60}^{\circ }=\frac{7}{2}$