Find the value of

tan \(\frac{\pi }{8}\) tan \(\frac{\pi }{12}\) tan \(\frac{3\pi }{8}\) tan \(\frac{5\pi }{12}\) - sin² \(\frac{\pi }{6}\)

\(\frac{3}{4}\)

⇒ tan A. tan B = 1 if A + B = 90°

Here,

\(\frac{\pi }{8}\) + \(\frac{3\pi }{8}\) = (\(\frac{\pi }{2}\)) = 90° and

\(\frac{\pi }{12}\) + \(\frac{5\pi }{12}\) = (\(\frac{\pi }{2}\)) = 90°

Therefore,

tan \(\frac{\pi }{8}\) tan \(\frac{\pi }{12}\) tan \(\frac{3\pi }{8}\) tan \(\frac{5\pi }{12}\) - sin² \(\frac{\pi }{6}\) = 1 × 1 - sin²\(\frac{\pi }{6}\)

= 1 - \(\frac{1}{4}\) = \(\frac{3}{4}\)