Question

Find the value of :
${3tan}^2{30}^{\circ }+\frac{4}{3}{cos}^2{30}^o-\frac{1}{2}{cot}^2{45}^o-\frac{2}{3}{sin}^2{60}^o+\frac{1}{8}{sec}^4{60}^o$

Answer 1

Substituting the values of $\tan {30}^{\circ }=\frac{1}{\sqrt{3}},\cos {30}^{\circ }=\frac{\sqrt{3}}{2},\cot {45}^{\circ }=1,\sin {60}^{\circ }=\frac{\sqrt{3}}{2},\sec {60}^{\circ }=2$

$=3{\left(\tan {30}^{\circ }\right)}^2+\frac{4}{3}{\left(\cos {30}^{\circ }\right)}^2-\frac{1}{2}{\left(\cot {45}^{\circ }\right)}^2-\frac{2}{3}{\left(\sin {60}^{\circ }\right)}^2+\frac{1}{8}{\left(\sec {60}^{\circ }\right)}^4$

$=3{\left(\frac{1}{\sqrt{3}}\right)}^2+\frac{4}{3}{\left(\frac{\sqrt{3}}{2}\right)}^2-\frac{1}{2}{\left(1\right)}^2-\frac{2}{3}{\left(\frac{\sqrt{3}}{2}\right)}^2+\frac{1}{8}{\left(2\right)}^4$

$=3\times \frac{1}{3}+\frac{4}{3}\times \frac{3}{4}-\frac{1}{2}-\frac{2}{3}\times \frac{3}{4}+\frac{16}{8}$

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

$=4-1=3$