The Value of \( \frac{1-\tan^2 15^\circ}{1+\tan^2 15^\circ} \)
Question
Find the value of
\[ \frac{1-\tan^2 15^\circ}{1+\tan^2 15^\circ} \]
(a) \(1\)
(b) \(\sqrt3\)
(c) \(\frac{\sqrt3}{2}\)
(d) \(2\)
Solution
Use the standard identity:
\[ \cos 2\theta = \frac{1-\tan^2\theta} {1+\tan^2\theta} \]
Putting
\[ \theta=15^\circ \]
we get
\[ \frac{1-\tan^2 15^\circ} {1+\tan^2 15^\circ} = \cos 30^\circ \]
Since
\[ \cos 30^\circ = \frac{\sqrt3}{2} \]
Therefore,
\[ \frac{1-\tan^2 15^\circ} {1+\tan^2 15^\circ} = \frac{\sqrt3}{2} \]
Final Answer
\[ \boxed{\frac{\sqrt3}{2}} \]
Hence, the correct option is (c) \(\frac{\sqrt3}{2}\).