Question

Question #e1576

Answer 1

`1.`

Explanation:

The Exp.`=cot(9/2)*cot(18/2)*cot(27/2)*...*cot(171/2).`

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^(st)` term and `19^(th)` term, i.e, `{cot(9/2)*cot(171/2)},`

Pair 2 : `2^(nd)` term and `18^(th)` term, i.e., `{cot(18/2)*cot(162/2)},`

`vdots vdots vdots vdots`

Pair 9: `9^(th)` term and `11^(th)`term, i.e.,{cot(81/2)*cot(99/2)},#

Recall that the `10^(th)` term, which is `cot(90/2),` is the one that

has remained unpaired.

`:." The Exp.="{cot(9/2)*cot(171/2)}{cot(18/2)*cot(162/2)}...{cot(81/2)*cot(99/2)}*cot(90/2),......(ast)`

Finally, we note that,

`alpha+beta=90 rArr cotalpha*cotbeta=cotalpha*cot(90-alpha),`

`=cotalpha*tanalpha=1.`

Hence, from `(ast)," the Exp.="{1}{1}...{1}[cot45]=1.`

Enjoy Maths.!

Answer 2

The Given Exp.

`=cot(9/2)*cot(18/2)*cot(27/2)*...*cot(171/2)`

`=prod_(r=1)^(r=19)cot((9r)/2)`

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

`=cot((9*10)/2)=cot45=1`

Now the product of 1st and 19th term

`cot(9/2)*cot(171/2)`

`=cot(9/2)*cot((180-9)/2)`

`=cot(9/2)*cot(180/2-9/2)`

`=cot(9/2)*cot(90-9/2)`

`=cot(9/2)*tan(9/2)=1`

Similarly product of 2nd 18th term

`=cot(18/2)*cot(162/2)`

`=cot(18/2)*cot((180-18)/2)`

`=cot(18/2)*cot(90-18/2)`

`=cot(18/2)*tan(18/2)=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((9r)/2)=1`