Question

The value of $\frac{1-{\tan }^2{15}^o}{1+{\tan }^2{15}^o}$ is equal to

• A. $1$
• B. $\sqrt{3}$
• C. $\frac{\sqrt{3}}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{\sqrt{3}}{2}$

We have,

$\frac{1-{\tan }^2{15}^o}{1+{\tan }^2{15}^o}\therefore \cos 2A=\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}$

$=\cos (2\times {15}^o)$

$=\cos {30}^o$

$=\frac{\sqrt{3}}{2}$

Hence, this is the answer.