Question

Find the value of $\frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}.$

• A. 0.5
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is A

0.5

$\text{We know that}tan{30}^{\circ }=\frac{1}{\sqrt{3}}.$

$\Rightarrow \frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}=\frac{1-(\frac{1}{\sqrt{3}}{)}^2}{1+(\frac{1}{\sqrt{3}}{)}^2}$

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

$=\frac{3-1}{3+1}$

$=\frac{2}{4}$ = $\frac{1}{2}=0.5$