Question

If $\theta =\frac{\pi }{4n}$ , then value of $\tan \theta \tan 2\theta ...\tan (2n-1)\theta$ equals

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

$1$

$\tan \left(\theta \right).\tan \left(2\theta \right)+...\tan \left(\right(2n-1\left)\theta \right)$.

$=\tan \left(\frac{\pi }{4n}\right).\tan \left(\frac{\pi }{2n}\right)..\tan (\frac{\pi }{2}-\frac{\pi }{4n})$.
$=\left[\tan \right(\frac{\pi }{4n}.\cot \left(\frac{\pi }{4n}\right)]\times [\tan \left(\frac{\pi }{2n}\right).\cot \left(\frac{\pi }{2n}\right)]\times [\tan \left(\frac{3\pi }{4n}\right).\cot \left(\frac{3\pi }{4n}\right)]...$
$=1$.