Find the real part of tan-11+i.
-tan-122
tan-122
-tan-1122
0
Simplifying the given function using Trigonometric identities
Let,
tan-11+i=a+ibtan-11-i=a-ib
Therefore,
tan-11+i+tan-11-i=a+ib+a-ib⇒tan-11+i+1-i1-(1+i)(1-i)=2a⇒tan-121-1+i2=2a⇒tan-1-2=2a⇒a=-tan-122
The real part of tan-11+i is a.
Therefore, Option(A) is the correct answer.