# Question e1576

Apr 10, 2017

$1.$

#### Explanation:

The Exp.$= \cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{18}{2}\right) \cdot \cot \left(\frac{27}{2}\right) \cdot \ldots \cdot \cot \left(\frac{171}{2}\right) .$

Observe that, there are $19$ terms in the Product.

If we make their pairs, there will be $9$ pairs, and, $1$ term will be

left ungrouped.

The pairs are constructed as under :

Pair 1 : ${1}^{s t}$ term and ${19}^{t h}$ term, i.e, $\left\{\cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{171}{2}\right)\right\} ,$

Pair 2 : ${2}^{n d}$ term and ${18}^{t h}$ term, i.e., $\left\{\cot \left(\frac{18}{2}\right) \cdot \cot \left(\frac{162}{2}\right)\right\} ,$

$\vdots \vdots \vdots \vdots$

Pair 9: ${9}^{t h}$ term and ${11}^{t h}$term, i.e.,{cot(81/2)*cot(99/2)},

Recall that the ${10}^{t h}$ term, which is $\cot \left(\frac{90}{2}\right) ,$ is the one that

has remained unpaired.

$\therefore \text{ The Exp.=} \left\{\cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{171}{2}\right)\right\} \left\{\cot \left(\frac{18}{2}\right) \cdot \cot \left(\frac{162}{2}\right)\right\} \ldots \left\{\cot \left(\frac{81}{2}\right) \cdot \cot \left(\frac{99}{2}\right)\right\} \cdot \cot \left(\frac{90}{2}\right) , \ldots \ldots \left(\ast\right)$

Finally, we note that,

$\alpha + \beta = 90 \Rightarrow \cot \alpha \cdot \cot \beta = \cot \alpha \cdot \cot \left(90 - \alpha\right) ,$

$= \cot \alpha \cdot \tan \alpha = 1.$

Hence, from $\left(\ast\right) , \text{ the Exp.=} \left\{1\right\} \left\{1\right\} \ldots \left\{1\right\} \left[\cot 45\right] = 1.$

Enjoy Maths.!

Apr 10, 2017

The Given Exp.

$= \cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{18}{2}\right) \cdot \cot \left(\frac{27}{2}\right) \cdot \ldots \cdot \cot \left(\frac{171}{2}\right)$

$= {\prod}_{r = 1}^{r = 19} \cot \left(\frac{9 r}{2}\right)$

There are 19 terms and 10th term is the middle term which is

$= \cot \left(\frac{9 \cdot 10}{2}\right) = \cot 45 = 1$

Now the product of 1st and 19th term

$\cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{171}{2}\right)$

$= \cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{180 - 9}{2}\right)$

$= \cot \left(\frac{9}{2}\right) \cdot \cot \left(\frac{180}{2} - \frac{9}{2}\right)$

$= \cot \left(\frac{9}{2}\right) \cdot \cot \left(90 - \frac{9}{2}\right)$

$= \cot \left(\frac{9}{2}\right) \cdot \tan \left(\frac{9}{2}\right) = 1$

Similarly product of 2nd 18th term

$= \cot \left(\frac{18}{2}\right) \cdot \cot \left(\frac{162}{2}\right)$

$= \cot \left(\frac{18}{2}\right) \cdot \cot \left(\frac{180 - 18}{2}\right)$

$= \cot \left(\frac{18}{2}\right) \cdot \cot \left(90 - \frac{18}{2}\right)$

$= \cot \left(\frac{18}{2}\right) \cdot \tan \left(\frac{18}{2}\right) = 1$

Similarly it can be shown that the product of each of 9 pairs (as paired in the manner above) will be $= 1$

Hence the given expression will have value $= 1$

i.e.

${\prod}_{r = 1}^{r = 19} \cot \left(\frac{9 r}{2}\right) = 1$