Question

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

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is B $1$

Using

$\sin x+\sin (x+y)+\sin (x+2y)+.....+\sin (x+(n-1\left)y\right)=\frac{\sin \left(\frac{ny}{2}\right)}{\sin \left(\frac{y}{2}\right)}\sin (x-\frac{n-1}{2}y)$

And

$\cos x+\cos (x+y)+\cos (x+2y)+.....+\cos (x+(n-1\left)y\right)=\frac{\sin \left(\frac{ny}{2}\right)}{\sin \left(\frac{y}{2}\right)}\cos (x-\frac{n-1}{2}y)$

We get,

$\tan \theta \tan 2\theta \tan 3\theta ....\tan (2n-2)\theta \tan (2n-1)\theta$

$=\frac{\sin \theta \sin 2\theta \sin 3\theta ....\sin (2n-2)\theta \sin (2n-1)\theta }{\cos \theta \cos 2\theta \cos 3\theta ...\cos (2n-2)\theta \cos (2n-1)\theta }$

$=\frac{\frac{\sin \left(\frac{(2n-1)\theta }{2}\right)}{\sin \left(\frac{\theta }{2}\right)}\sin (\theta +\frac{2n-2}{2}\theta )}{\frac{\sin \left(\frac{(2n-1)\theta }{2}\right)}{\sin \left(\frac{\theta }{2}\right)}\cos (\theta +\frac{2n-2}{2}\theta )}$

$=\tan \left(n\theta \right)=\tan \left(n\frac{\pi }{4n}\right)=1$