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Question Number 189998 by 073 last updated on 25/Mar/23

$$\mathrm{tanA}=\frac{1}{\mathrm{n}+1}$$$$\mathrm{tanB}=\frac{\mathrm{n}}{2\mathrm{n}+1}$$$$\Rightarrow \mathrm{A}+\mathrm{B}=?$$

Answered by Frix last updated on 25/Mar/23

$$A={\tan }^{-1}\frac{1}{n+1}$$$$B={\tan }^{-1}\frac{n}{2n+1}$$$$\tan ({\tan }^{-1}x+{\tan }^{-1}y)=\frac{x+y}{1-xy}$$$$\tan (A+B)=\frac{n^2+3n+1}{2n^2+2n+1}$$

Commented by 073 last updated on 25/Mar/23

$$\mathrm{only}\mathrm{A}+\mathrm{B}=??$$

Commented by Frix last updated on 25/Mar/23

$$\mathrm{Sorry}\mathrm{but}\mathrm{this}\mathrm{is}\mathrm{a}\mathrm{stupid}\mathrm{question}.$$$$A+B={\tan }^{-1}\frac{n^2+3n+1}{2n^2+2n+1}$$

Commented by Spillover last updated on 27/Mar/23

$$\mathrm{of}\mathrm{course}.\mathrm{it}\mathrm{is}\mathrm{stupid}\mathrm{question}$$

Answered by Spillover last updated on 01/Jul/23

$$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$$$=\frac{\frac{1}{1+n}+\frac{n}{2n+1}}{1-\left[\frac{1}{1+n}\times \frac{n}{2n+1}\right]}$$$$\tan (A+B)=\frac{3n^2+3n+1}{3n^2+3n+1}$$$$A+B={\tan }^{-1}1$$$$A+B=\frac{\pi }{4}$$

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