Question

If $4n\alpha =\pi$, then the value of $\cot \alpha \cdot \cot 2\alpha \cdot \cot 3\alpha {\cdot }^{\prime }...{.}^{\prime }\cot (2n-1)\alpha$ is

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is B $1$

we can write $n=\frac{\pi }{4\alpha }$

Now, $\cot (2n-1)\alpha =\cot (\frac{2\pi }{4\alpha }-1)\alpha =\cot (\frac{\pi }{2}-\alpha )=\tan \alpha$

Similarly, $\cot (2n-2)\alpha =\cot (\frac{2\pi }{4\alpha }-2)\alpha =\cot (\frac{\pi }{2}-2\alpha )=\tan 2\alpha$ and so on

So, $\cot \alpha .\cot 2\alpha ..............\cot (2n-2)\alpha .\cot (2n-1)\alpha =\cot \alpha .\cot 2\alpha ...........\tan 2\alpha \tan \alpha =\left(\cot \alpha \tan \alpha \right)\left(\cot 2\alpha \tan 2\alpha \right)\left(\cot 3\alpha \tan 3\alpha \right).......=\mathrm{1.1.1...............}=1$

Therefore, Correct Answer is $1$