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Question Number 116114 by Khalmohmmad last updated on 01/Oct/20

Answered by Dwaipayan Shikari last updated on 01/Oct/20

$${\tan }^{-1}(3+2\sqrt{2})-{\tan }^{-1}\left(\frac{1}{\sqrt{2}}\right)$$$${\tan }^{-1}\frac{3+2\sqrt{2}-\frac{1}{\sqrt{2}}}{1+(3+2\sqrt{2})\frac{1}{\sqrt{2}}}={\tan }^{-1}\left(\frac{3+\frac{3}{\sqrt{2}}}{1+\frac{3}{\sqrt{2}}+2}\right)={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$$

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