Maurice Tutor

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    Argosy University/ Phoniex University/
    Nov-2005 - Oct-2011

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    Phoniex University
    Oct-2001 - Nov-2016

Category > Management Posted 10 Jul 2017 My Price 8.00

tan(Π-x)=-tan(x); --(1)

I can't find where I went wrong!! Can anybody please find it for me?

We already have:

tan(Π-x)=-tan(x); --(1)

and

tan(-x)=-tan(x); --(2)

So we start:

Let x = tan-1(-y),

Then -y = tan(x);

=> y = -tan(x);

=> y = tan(Π-X) and y = tan(-x) using (1) and (2)

=> Π-x=tan-1(y) and -x= tan-1(y)

=> -x = tan-1(y)-Π and -x = tan-1(y)

=> x = Π - tan-1(y) and x = -tan-1(y)

=> Π - tan-1(y) = - tan-1(y)

=> Π = 0

How can it be possible? Where I went wrong??

Please help me!

Answers

Maurice Tutor
(5)
Status NEW Posted 10 Jul 2017 02:07 PM My Price 8.00

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