Tan⁡[Tan−1⁡(1a+b)+Tan−1⁡(ba2+ab+1)]=

Tan⁡[Tan−1⁡(1a+b)+Tan−1⁡((a+b)−a1+(a+b)a)]=Tan⁡[Tan−1⁡(1a+b)+Tan−1⁡(a+b)−Tan−1⁡a]=1a

Tan⁡[Tan−1⁡(1a+b)+Tan−1⁡((a+b)−a1+(a+b)a)]=Tan⁡[Tan−1⁡(1a+b)+Tan−1⁡(a+b)−Tan−1⁡a]=1a

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Offline Centres

$Tan [{Tan}^{- 1} (\frac{1}{a + b}) + {Tan}^{- 1} (\frac{b}{a^{2} + a b + 1})] =$

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

1/a

1/b

answer is B.

(Unlock A.I Detailed Solution for FREE)

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

$Tan [{Tan}^{- 1} (\frac{1}{a + b}) + {Tan}^{- 1} (\frac{(a + b) - a}{1 + (a + b) a})]$

$= Tan [{Tan}^{- 1} (\frac{1}{a + b}) + {Tan}^{- 1} (a + b) - {Tan}^{- 1} a] = \frac{1}{a}$

Watch 3-min video & get full concept clarity

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test