Question

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

• A. 2
• B. 0.5
• C. 4
• D. 1

Answer 1

The correct option is B

0.5

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

Now, putting in the value of $tan{30}^{\circ }$ in the provided expression.

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

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

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

$\Rightarrow \frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}=\frac{2}{4}$

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

$\Rightarrow \frac{1-ta{n}^2{30}^{\circ }}{1+ta{n}^2{30}^{\circ }}=0.5$