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!

Question

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!

SAGAR SINGH - IIT DELHI · Answer

Dear student, Please go thru Inverse trignometric functions, There are some coditions when inverse of a function is applied.....

Davis Devasia · Answer

Can you please note it here? I couldn't find where I am wrong!

2 Answers