Question

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

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

$1$

There are $(2n-1)$ factors;

Of these let us group a pair of products, selecting the corresponding parts from either end; thus one middle will be left out without any pair for the same.

So the arrangement for them is:

$P=\left[cot\right(2na-a)\ast cot(a\left)\right]\ast \left[cot\right(2na-2a)\ast cot(2a\left)\right]\ast \left[cot\right(2na-3a)\ast cot(3a\left)\right]\ast .....\ast cot\left(na\right)]..........(3)$

When $4na=\pi ,na=\pi /4;$

So

$cot\left(na\right)=cot\left(\pi /4\right)=cot(2na-a)=cot(\pi /2-a)=tan\left(a\right)$

So

$cot(2na-a)cot\left(a\right)=tan\left(a\right)cot\left(a\right)=1$

Similarly

$cot(2na-2a)cot\left(2a\right)=cot(\pi /2-2a)cot\left(2a\right)=tan\left(2a\right)cot\left(2a\right)=1$

Similarly every pair will simplifies to $1$

Substituting all these in $eq\left(3\right)$. above, $P=1\times 1\times 1\times 1\times 1\times ....................1\times 1=1$

Thus the answer is $1.$