Prove that:\(tan^{-1}2-tan^{-1}1=tan^{-1}\frac{1}{3}\)

1.	Prove that:\(tan^{-1}2-tan^{-1}1=tan^{-1}\frac{1}{3}\)
Answer» To Prove : \(tan^{-1}2-tan^{-1}1=tan^{-1}\frac{1}{3}\) Formula Used : \(tan^{-1}x-tan^{-1}y= tan^{-1}\left(\frac{x+y}{1+xy}\right)\) where xy > -1 Proof: LHS = tan^-12 – tan^-1 1 \(=tan^{-1}\left(\frac{2-1}{1+2}\right)\) \(=tan^{-1}\left(\frac{1}{3}\right)\) = RHS Therefore LHS = RHS Hence proved.

Answer»

To Prove : \(tan^{-1}2-tan^{-1}1=tan^{-1}\frac{1}{3}\)

Formula Used :

\(tan^{-1}x-tan^{-1}y= tan^{-1}\left(\frac{x+y}{1+xy}\right)\) where xy > -1

Proof:

LHS = tan^-12 – tan^-1 1

\(=tan^{-1}\left(\frac{2-1}{1+2}\right)\)

\(=tan^{-1}\left(\frac{1}{3}\right)\)

= RHS

Therefore LHS = RHS

Hence proved.

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