Question

${\cos }^2{30}^{\circ }+si{n}^2{45}^{\circ }-\frac{1}{3}ta{n}^2{60}^{\circ }=$

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{3}{4}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D

$\frac{1}{4}$

${\cos }^2{30}^{\circ }+si{n}^2{45}^{\circ }-\frac{1}{3}ta{n}^2{60}^{\circ }$

$=(cos{30}^{\circ }{)}^2+(sin{45}^{\circ }{)}^2-\frac{1}{3}(tan{60}^{\circ }{)}^2$

(We know that,
$cos{30}^{\circ }=\frac{\sqrt{3}}{2}$, $sin{45}^{\circ }=\frac{1}{\sqrt{2}}$ and $tan{60}^{\circ }=\sqrt{3})$

$=(\frac{\sqrt{3}}{2}{)}^2+(\frac{1}{\sqrt{2}}{)}^2-\frac{1}{3}(\sqrt{3}{)}^2$

$=\frac{3}{4}+\frac{1}{2}-\frac{3}{3}$

$=\frac{3+2}{4}-1$

$=\frac{1}{4}$