Question

Evaluate : $\frac{2}{3}({\cos }^4{30}^o-{\sin }^4{45}^o)-3({\sin }^2{60}^o-{\sec }^2{45}^o)+\frac{1}{4}{\cot }^2{30}^o$

• A. $\frac{163}{24}$
• B. $\frac{123}{24}$
• C. $\frac{113}{24}$
• D. $\frac{183}{24}$

Answer 1

The correct option is C $\frac{113}{24}$

Given

Now,

$\frac{2}{3}({\cos }^4{30}^0-{\sin }^4{45}^0)$

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

$=\frac{2}{3}\left(\frac{9-4}{16}\right)$

$⟹\frac{5}{24}$

Now,

$3({\sin }^2{60}^0-{\sec }^2{45}^0)$

$=3({\left(\frac{\sqrt{3}}{2}\right)}^2-(\sqrt{2}{)}^2)$

$=3(\frac{3}{4}-2)$

$=3\left(\frac{-5}{4}\right)$

$\left(\frac{-15}{4}\right)$

Now,

$=\frac{1}{4}{\cot }^2{30}^0$

$=\frac{1}{4}(\sqrt{3}{)}^2$

$=\frac{3}{4}$

$\frac{2}{3}({\cos }^{4}30-{\sin }^{4}45)+3({\sin }^{2}60-{\sec }^{2}45)+\frac{1}{4}{\cot }^{2}30$

$=\frac{5}{24}+\frac{15}{4}+\frac{3}{4}$

$⟹\frac{5+90+18}{24}$

$⟹\frac{113}{24}$