Question

The value of $2co{t}^2\left(\frac{\mathrm{\pi }}{6}\right)+4{\tan }^2\left(\frac{\mathrm{\pi }}{6}\right)-3\cos ec\left(\frac{\mathrm{\pi }}{6}\right)$ is

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

Answer 1

The correct option is C

$\frac{4}{3}$

Explanation for the correct answer:

$\tan \left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{\sqrt{3}}$

$\Rightarrow$ $cot\left(\frac{\mathrm{\pi }}{6}\right)=\frac{1}{\tan \left({\displaystyle \frac{\mathrm{\pi }}{6}}\right)}=\sqrt{3}$

$1+co{t}^2\left(\theta \right)=\cos e{c}^2\left(\theta \right)$

$\Rightarrow$ $1+{\left(\sqrt{3}\right)}^2=\cos e{c}^2\left(\frac{\mathrm{\pi }}{6}\right)$

$\Rightarrow$ $\cos e{c}^2\left(\frac{\mathrm{\pi }}{6}\right)=4$

$\Rightarrow$ $\cos ec\left(\frac{\mathrm{\pi }}{6}\right)=2$

The given expression is $2co{t}^2\left(\frac{\mathrm{\pi }}{6}\right)+4{\tan }^2\left(\frac{\mathrm{\pi }}{6}\right)-3\cos ec\left(\frac{\mathrm{\pi }}{6}\right)$

Substituting all the values, we get

$2co{t}^2\left(\frac{\mathrm{\pi }}{6}\right)+4{\tan }^2\left(\frac{\mathrm{\pi }}{6}\right)-3\cos ec\left(\frac{\mathrm{\pi }}{6}\right)=2{\left(\sqrt{3}\right)}^2+4\frac{1}{{\left(\sqrt{3}\right)}^2}-3\left(2\right)$

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

$\Rightarrow$ $2co{t}^2\left(\frac{\mathrm{\pi }}{6}\right)+4{\tan }^2\left(\frac{\mathrm{\pi }}{6}\right)-3\cos ec\left(\frac{\mathrm{\pi }}{6}\right)=\frac{4}{3}$

Hence, the value of the expression $2co{t}^2\left(\frac{\mathrm{\pi }}{6}\right)+4{\tan }^2\left(\frac{\mathrm{\pi }}{6}\right)-3\cos ec\left(\frac{\mathrm{\pi }}{6}\right)$ is $\frac{4}{3}$

Hence, option (C) is the correct answer.