If tan A = 1 x x 2 + x + 1 , tan B = x x 2 + x + 1 and tan C = x − 3 + x − 2 + x − 1 1 2 , 0 < A , B , C…

If tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112,0<A,B,C<π2, then A+B is equal to:
  1. C
  2. π−C
  3. 2π−C
  4. π2−C

Solution

Given, tanA=1xx2+x+1,tanB=xx2+x+1 and tanC=x−3+x−2+x−112=1+x+x2xx

We know that,

tanA+B=tanA+tanB1−tanAtanB

⇒tanA+B=1xx2+x+1+xx2+x+11−1x2+x+1

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=1+xx2+x+1x2+xx

⇒tanA+B=x2+x+1xx

⇒tanA+B=tanC

⇒A+B=C

Asked in: JEE Main 2024 (01 Feb Shift 1)

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