# (1+tan 1)(1+tan 2).....(1+tan 45)=

• Oct 21st 2011, 09:35 PM
Raj917
(1+tan 1)(1+tan 2).....(1+tan 45)=
(1+tan 1)(1+tan 2).....(1+tan 45)=

What should i be knowing to find the answer to this ?
• Oct 22nd 2011, 12:13 AM
Opalg
Re: (1+tan 1)(1+tan 2).....(1+tan 45)=
Quote:

Originally Posted by Raj917
(1+tan 1)(1+tan 2).....(1+tan 45)=

What should i be knowing to find the answer to this ?

First, you can write it as \$\displaystyle (1+\tan 0^\circ)(1+\tan 1^\circ)(1+\tan 2^\circ)\cdots(1+\tan 45^\circ)\$ (because the additional term at the beginning is equal to 1).

Next, pair off the terms from the beginning and the end:

\$\displaystyle \bigl((1+\tan 0^\circ)(1+\tan 45^\circ)\bigr) \bigl((1+\tan 1^\circ)(1+\tan 44^\circ)\bigr)\cdots\bigl((1+\tan 22^\circ)(1+\tan 23^\circ)\bigr).\$

Then see if you can work out the value of each of those paired products.
• Oct 22nd 2011, 06:48 AM
Raj917
Re: (1+tan 1)(1+tan 2).....(1+tan 45)=
Quote:

Originally Posted by Opalg
First, you can write it as \$\displaystyle (1+\tan 0^\circ)(1+\tan 1^\circ)(1+\tan 2^\circ)\cdots(1+\tan 45^\circ)\$ (because the additional term at the beginning is equal to 1).

Next, pair off the terms from the beginning and the end:

\$\displaystyle \bigl((1+\tan 0^\circ)(1+\tan 45^\circ)\bigr) \bigl((1+\tan 1^\circ)(1+\tan 44^\circ)\bigr)\cdots\bigl((1+\tan 22^\circ)(1+\tan 23^\circ)\bigr).\$

Then see if you can work out the value of each of those paired products.

Thank you sir the answer is 2^23