Question

If $4n\alpha =\pi$ then the value of $\tan \alpha \tan 2\alpha \tan 3\alpha ...\tan (2n-1)\alpha =$

• A. $1$
• B. $0$
• C. $-1$
• D.
  none of these

Answer 1

The correct option is A $1$

According to question,

$4n\alpha =\pi$

=> $2n\alpha =\frac{\pi }{2}$

And , $(2n-1)\alpha =2n\alpha -\alpha =\frac{\pi }{2}-\alpha$

Or, $tan\alpha tan(2n-1)\alpha =1$

=> $tan(2n-1)\alpha =cot\alpha$

Similarly,

=> $tan(2n-2)\alpha =cot2\alpha$ and so on.....

Combine terms equidistant from the beginning and end , and each product is $1$.