Question

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

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

Answer 1

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

$\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\therefore \frac{1-{\tan }^{2}15}{1+{\tan }^{2}15}=\cos (2\times 15)$

$\cos \left({30}^o\right)=\frac{\sqrt{3}}{2}$