Question #e1576

2 Answers
Apr 10, 2017

#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.!

Apr 10, 2017

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#