Question

Evaluate : $\frac{2{\cos }^2{60}^{\circ }+3{\sec }^2{30}^{\circ }-2{\tan }^2{45}^{\circ }}{{\sin }^2{30}^{\circ }+{\cos }^2{45}^{\circ }}$.

Answer 1

We know,

$\sin {30}^o=\frac{1}{2}$ and $\sec {30}^o=\frac{2}{\sqrt{3}}$

$\cos {45}^o=\frac{1}{\sqrt{2}}$ and $\tan {45}^o=1$

$\cos {60}^o=\frac{1}{2}$

So, $\frac{2{\cos }^2{60}^{\circ }+3{\sec }^2{30}^{\circ }-2{\tan }^2{45}^{\circ }}{{\sin }^2{30}^{\circ }+{\cos }^2{45}^{\circ }}$$=\frac{2{\left(\frac{1}{2}\right)}^2+3{\left(\frac{2}{\sqrt{3}}\right)}^2-2(1{)}^2}{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}$

$=\frac{2\left(\frac{1}{4}\right)+3\left(\frac{4}{3}\right)-2}{\frac{1}{4}+\frac{1}{2}}$

$=\frac{\frac{1}{2}+4-2}{\frac{3}{4}}=\frac{5}{2}\times \frac{4}{3}$

$=\frac{10}{3}$