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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!

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!

Grade:12

2 Answers

SAGAR SINGH - IIT DELHI
878 Points
13 years ago

Dear student,

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

Davis Devasia
31 Points
13 years ago

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

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