Question

The value of $\frac{3}{4}{\tan }^2{30}^{\circ }-3{\sin }^2{60}^{\circ }+cose{c}^2{45}^{\circ }$ is

• A. $1$
• B. $8$
• C. $0$
• D. $12$

Answer 1

The correct option is C $0$

Substituting the corresponding trigonometric values, we get
$\frac{3}{4}(\frac{1}{\sqrt{3}}{)}^2-3(\frac{\sqrt{3}}{2}{)}^2+(\sqrt{2}{)}^2$
$=\frac{3}{4}.\frac{1}{3}-3.\frac{3}{4}+2$
$=\frac{1}{4}-\frac{9}{4}+2$
$=\frac{-8}{4}+2$
$=-2+2$
$=0$