tanA-1-n-1-tanB-n-2n-1-A-B-

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

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

Answered by Frix last updated on 25/Mar/23

A=tan^(−1) (1/(n+1)) B=tan^(−1) (n/(2n+1)) tan (tan^(−1) x +tan^(−1) y) =((x+y)/(1−xy)) tan (A+B)=((n^2 +3n+1)/(2n^2 +2n+1))

$${A}=\mathrm{tan}^{−\mathrm{1}} \:\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$${B}=\mathrm{tan}^{−\mathrm{1}} \:\frac{{n}}{\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{tan}\:\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\:+\mathrm{tan}^{−\mathrm{1}} \:{y}\right)\:=\frac{{x}+{y}}{\mathrm{1}−{xy}} \\ $$$$\mathrm{tan}\:\left({A}+{B}\right)=\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}} \\ $$

Commented by 073 last updated on 25/Mar/23

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

Commented by Frix last updated on 25/Mar/23

Sorry but this is a stupid question. A+B=tan^(−1) ((n^2 +3n+1)/(2n^2 +2n+1))

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

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)=((tan A+tan B)/(1−tan Atan B)) =(((1/(1+n))+(n/(2n+1)))/(1−[(1/(1+n))×(n/(2n+1))])) tan (A+B)=((3n^2 +3n+1)/(3n^2 +3n+1)) A+B=tan^(−1) 1 A+B=(π/4)

$$\mathrm{tan}\:\left({A}+{B}\right)=\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{B}}{\mathrm{1}−\mathrm{tan}\:{A}\mathrm{tan}\:{B}} \\ $$$$=\frac{\frac{\mathrm{1}}{\mathrm{1}+{n}}+\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}}{\mathrm{1}−\left[\frac{\mathrm{1}}{\mathrm{1}+{n}}×\frac{{n}}{\mathrm{2}{n}+\mathrm{1}}\right]} \\ $$$$\mathrm{tan}\:\left({A}+{B}\right)=\frac{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}}{\mathrm{3}{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{1}} \\ $$$${A}+{B}=\mathrm{tan}^{−\mathrm{1}} \mathrm{1} \\ $$$${A}+{B}=\frac{\pi}{\mathrm{4}} \\ $$

Facebook Tweet Pin

tanA-1-n-1-tanB-n-2n-1-A-B-

Leave a Reply Cancel reply