Question

The value of x from the equation $\frac{x}{\cos {45}^{\circ }.\sin {60}^{\circ }}={\tan }^2-{\tan }^2{30}^{\circ }$ is

• A. $3$
• B. $2$
• C. $1$
• D. $\frac{4}{\sqrt{6}}$

Answer 1

The correct option is D $\frac{4}{\sqrt{6}}$

Consider the given expression,

$\frac{x}{\cos {45}^0.\sin {60}^0}={\tan }^2{60}^0-{\tan }^2{30}^0$

$x=\cos {45}^0.\sin {60}^0({\tan }^2{60}^0-{\tan }^2{30}^0)$

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

$=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}(3-\frac{1}{3})$

$=\frac{\sqrt{3}}{2\sqrt{2}}\left(\frac{8}{3}\right)$

$=\frac{4}{\sqrt{6}}$

Hence, this is the answer.