Question

The value of $\frac{1-{\tan }^{2}15°}{1+{\tan }^{2}15°}$ is
(a) 1

(b) $\sqrt{3}$

(c) $\frac{\sqrt{3}}{2}$

(d) 2

Answer 1

$$\begin{gathered} \frac{1-{\tan }^{2}15°}{1+{\tan }^{2}15°} \\ \left[\mathrm{Using}\mathrm{identity}:\frac{1-{\tan }^{2}A}{1+{\tan }^{2}A}=\cos 2A\right] \end{gathered}$$
$\begin{array}{rcl}\mathrm{W}e\mathrm{have},\frac{1-{\tan }^{2}15°}{1+{\tan }^{2}15°}& =& \cos 2\left(15°\right)\\ & =& \cos 30°\\ & =& \frac{\sqrt{3}}{2}\end{array}$
Hence, the correct answer is option C.