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

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

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

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

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

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