Question

If $3{\sin }^2{30}^o+2{\tan }^2{60}^{\circ }-5{\cos }^2{45}^{\circ }$ is $4\frac{m}{4}$ then, $m$ is

Answer 1

$3{\sin }^2{30}^{\circ }+2{\tan }^2{60}^{\circ }-5{\cos }^2{45}^{\circ }$
= $3(\frac{1}{2}{)}^2+2(\sqrt{3}{)}^2-5(\frac{1}{\sqrt{2}}{)}^2$

= $\frac{3}{4}+6-\frac{5}{2}$

= $\frac{3+24-10}{4}$

= $\frac{17}{4}$

= $4\frac{1}{4}$