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Question Number 82929 by VBash last updated on 26/Feb/20

if 2B+A=45° show that; tan B= ((1−2tanA−tan^2 A)/(1+2tanA−tan^2 A))

$${if}\:\mathrm{2}{B}+{A}=\mathrm{45}° \\ $$$${show}\:{that}; \\ $$$${tan}\:{B}=\:\frac{\mathrm{1}−\mathrm{2}{tanA}−{tan}^{\mathrm{2}} {A}}{\mathrm{1}+\mathrm{2}{tanA}−{tan}^{\mathrm{2}} {A}} \\ $$

Answered by jagoll last updated on 26/Feb/20

A+B = 45^o − B tan (A+B) = tan (45^o −B) ((tan A+tan B)/(1−tan A.tan B)) = ((1−tan B)/(1+tan B)) (tan A+tan B)(1+tan B) = (1−tanA tan B)(1−tan B) tan A = ((tan^2 B+2tan B−1)/(tan^2 B−2tan B−1)) or tan A = ((1−2tan B−tan^2 B)/(1+2tan B−tan^2 B))

$$\mathrm{A}+\mathrm{B}\:=\:\mathrm{45}^{\mathrm{o}} \:−\:\mathrm{B} \\ $$$$\mathrm{tan}\:\left(\mathrm{A}+\mathrm{B}\right)\:=\:\mathrm{tan}\:\left(\mathrm{45}^{\mathrm{o}} −\mathrm{B}\right) \\ $$$$\frac{\mathrm{tan}\:\mathrm{A}+\mathrm{tan}\:\mathrm{B}}{\mathrm{1}−\mathrm{tan}\:\mathrm{A}.\mathrm{tan}\:\mathrm{B}}\:=\:\frac{\mathrm{1}−\mathrm{tan}\:\mathrm{B}}{\mathrm{1}+\mathrm{tan}\:\mathrm{B}} \\ $$$$\left(\mathrm{tan}\:\mathrm{A}+\mathrm{tan}\:\mathrm{B}\right)\left(\mathrm{1}+\mathrm{tan}\:\mathrm{B}\right)\:=\: \\ $$$$\left(\mathrm{1}−\mathrm{tanA}\:\mathrm{tan}\:\mathrm{B}\right)\left(\mathrm{1}−\mathrm{tan}\:\mathrm{B}\right) \\ $$$$\mathrm{tan}\:\mathrm{A}\:=\:\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{B}+\mathrm{2tan}\:\mathrm{B}−\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{B}−\mathrm{2tan}\:\mathrm{B}−\mathrm{1}} \\ $$$$\mathrm{or}\:\mathrm{tan}\:\mathrm{A}\:=\:\frac{\mathrm{1}−\mathrm{2tan}\:\mathrm{B}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{B}}{\mathrm{1}+\mathrm{2tan}\:\mathrm{B}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{B}} \\ $$$$ \\ $$

Commented by john santu last updated on 26/Feb/20

good sir

$$\mathrm{good}\:\mathrm{sir} \\ $$

Commented by som(math1967) last updated on 26/Feb/20

How tan(45°−B)=((1−tanA)/(1+tanA)) ???

$${How}\:{tan}\left(\mathrm{45}°−{B}\right)=\frac{\mathrm{1}−{tanA}}{\mathrm{1}+{tanA}}\:??? \\ $$

Commented by jagoll last updated on 26/Feb/20

hahaha..sorry. it my typo

$$\mathrm{hahaha}..\mathrm{sorry}.\:\mathrm{it}\:\mathrm{my}\:\mathrm{typo} \\ $$

Answered by TANMAY PANACEA last updated on 26/Feb/20

tan(2B)=tan(45−A) let tanA=a tanB=b ((2b)/(1−b^2 ))=((1−a)/(1+a)) ((2b+1−b^2 )/(2b−1+b^2 ))=((1−a+1+a)/(1−a−1−a)) ((2b+1−b^2 )/(2b−1+b^2 ))=(2/(−2a)) ((−a)/1)=((2b−1+b^2 )/(2b+1−b^2 )) a=((b^2 +2b−1)/(b^2 −2b−1)) pls recheck the question

$${tan}\left(\mathrm{2}{B}\right)={tan}\left(\mathrm{45}−{A}\right) \\ $$$${let}\:{tanA}={a}\:\:\:{tanB}={b} \\ $$$$\frac{\mathrm{2}{b}}{\mathrm{1}−{b}^{\mathrm{2}} }=\frac{\mathrm{1}−{a}}{\mathrm{1}+{a}} \\ $$$$\frac{\mathrm{2}{b}+\mathrm{1}−{b}^{\mathrm{2}} }{\mathrm{2}{b}−\mathrm{1}+{b}^{\mathrm{2}} }=\frac{\mathrm{1}−{a}+\mathrm{1}+{a}}{\mathrm{1}−{a}−\mathrm{1}−{a}} \\ $$$$\frac{\mathrm{2}{b}+\mathrm{1}−{b}^{\mathrm{2}} }{\mathrm{2}{b}−\mathrm{1}+{b}^{\mathrm{2}} }=\frac{\mathrm{2}}{−\mathrm{2}{a}} \\ $$$$\frac{−{a}}{\mathrm{1}}=\frac{\mathrm{2}{b}−\mathrm{1}+{b}^{\mathrm{2}} }{\mathrm{2}{b}+\mathrm{1}−{b}^{\mathrm{2}} } \\ $$$${a}=\frac{{b}^{\mathrm{2}} +\mathrm{2}{b}−\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{2}{b}−\mathrm{1}}\:\:{pls}\:{recheck}\:{the}\:{question} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by jagoll last updated on 26/Feb/20

it that the question wrong

$$\mathrm{it}\:\mathrm{that}\:\mathrm{the}\:\mathrm{question}\:\mathrm{wrong} \\ $$

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