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

$\frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}$

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

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

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