Question

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

• A. $\frac{7}{2}$
• B. $\frac{2}{7}$
• C. $\frac{17}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{7}{2}$

$tan{45}^{\circ }=1,cos{30}^{\circ }=\frac{\sqrt{3}}{2}andsin{60}^{\circ }=\frac{\sqrt{3}}{2}\Rightarrow 2ta{n}^2{45}^{\circ }+3co{s}^2{30}^{\circ }-si{n}^2{60}^{\circ }=2(1{)}^2+3(\frac{\sqrt{3}}{2}{)}^2-(\frac{\sqrt{3}}{2}{)}^2=2+\frac{9}{4}-\frac{3}{4}=\frac{8+9-3}{4}=\frac{14}{4}=\frac{7}{2}$