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Question Number 204083 by Tawa11 last updated on 05/Feb/24

Given that tan(A + B) = 1 and tan(A - B) = 1/7 find tan A and tan B

Given that tan(A + B) = 1 and tan(A - B) = 1/7 find tan A and tan B

Answered by a.lgnaoui last updated on 06/Feb/24

posons t1=tan A t2=tan B { ((((t1+t2)/(1−t1t2))= 1)),((((t1−t2)/(1+t1t2))=(1/7))) :} t1+t2=1−t1t2 t1−t2=(1/7)+(1/7)t1t2 2t1=((8−6t1t2)/7) t1=(4/(3t2+7)) t2+(4/(3t2+7))=1−((4t2)/(3t2+7)) 3(t2)^2 +8t2−3=0 △′=5^2 { ((t2=(1/6))),((t2=((−3)/2))) :} a) t2=(1/6) t1=(8/(15)) b)t2=((−3)/2) t1=−(8/(23)) tan A=(8/(15)) tan B=(1/6) tan A= (8/(23)) tan B=((−3)/2)

$$\boldsymbol{\mathrm{posons}}\:\:\boldsymbol{\mathrm{t}}\mathrm{1}=\mathrm{tan}\:\boldsymbol{\mathrm{A}}\:\:\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{2}=\mathrm{tan}\:\boldsymbol{\mathrm{B}} \\ $$$$\begin{cases}{\frac{\boldsymbol{\mathrm{t}}\mathrm{1}+\boldsymbol{\mathrm{t}}\mathrm{2}}{\mathrm{1}−\boldsymbol{\mathrm{t}}\mathrm{1}\boldsymbol{\mathrm{t}}\mathrm{2}}=\:\mathrm{1}}\\{\frac{\boldsymbol{\mathrm{t}}\mathrm{1}−\boldsymbol{\mathrm{t}}\mathrm{2}}{\mathrm{1}+\boldsymbol{\mathrm{t}}\mathrm{1}\boldsymbol{\mathrm{t}}\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{7}}}\end{cases} \\ $$$$\boldsymbol{\mathrm{t}}\mathrm{1}+\boldsymbol{\mathrm{t}}\mathrm{2}=\mathrm{1}−\boldsymbol{\mathrm{t}}\mathrm{1}\boldsymbol{\mathrm{t}}\mathrm{2} \\ $$$$\boldsymbol{\mathrm{t}}\mathrm{1}−\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{\mathrm{1}}{\mathrm{7}}+\frac{\mathrm{1}}{\mathrm{7}}\boldsymbol{\mathrm{t}}\mathrm{1}\boldsymbol{\mathrm{t}}\mathrm{2} \\ $$$$\mathrm{2}\boldsymbol{\mathrm{t}}\mathrm{1}=\frac{\mathrm{8}−\mathrm{6}\boldsymbol{\mathrm{t}}\mathrm{1}\boldsymbol{\mathrm{t}}\mathrm{2}}{\mathrm{7}}\:\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{1}=\frac{\mathrm{4}}{\mathrm{3}\boldsymbol{\mathrm{t}}\mathrm{2}+\mathrm{7}} \\ $$$$\boldsymbol{\mathrm{t}}\mathrm{2}+\frac{\mathrm{4}}{\mathrm{3}\boldsymbol{\mathrm{t}}\mathrm{2}+\mathrm{7}}=\mathrm{1}−\frac{\mathrm{4}\boldsymbol{\mathrm{t}}\mathrm{2}}{\mathrm{3}\boldsymbol{\mathrm{t}}\mathrm{2}+\mathrm{7}} \\ $$$$\mathrm{3}\left(\boldsymbol{\mathrm{t}}\mathrm{2}\right)^{\mathrm{2}} +\mathrm{8}\boldsymbol{\mathrm{t}}\mathrm{2}−\mathrm{3}=\mathrm{0} \\ $$$$\:\:\bigtriangleup'=\mathrm{5}^{\mathrm{2}} \:\:\:\:\:\begin{cases}{\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{\mathrm{1}}{\mathrm{6}}}\\{\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{−\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$$\left.\boldsymbol{\mathrm{a}}\right)\:\:\:\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{\mathrm{1}}{\mathrm{6}}\:\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{1}=\frac{\mathrm{8}}{\mathrm{15}} \\ $$$$ \\ $$$$ \\ $$$$\left.\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{t}}\mathrm{2}=\frac{−\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\boldsymbol{\mathrm{t}}\mathrm{1}=−\frac{\mathrm{8}}{\mathrm{23}} \\ $$$$ \\ $$$$\:\:\: \\ $$$$\mathrm{tan}\:\mathrm{A}=\frac{\mathrm{8}}{\mathrm{15}}\:\:\:\:\:\:\:\:\mathrm{tan}\:\mathrm{B}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\mathrm{tan}\:\mathrm{A}=\:\frac{\mathrm{8}}{\mathrm{23}}\:\:\:\:\:\:\:\:\mathrm{tan}\:\mathrm{B}=\frac{−\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\: \\ $$$$\: \\ $$

Commented by Frix last updated on 06/Feb/24

Method ok but calculation error: 3t_2 ^2 +8t_2 −3=0 ⇒ t_2 =−3∨t_2 =(1/3)

$$\mathrm{Method}\:\mathrm{ok}\:\mathrm{but}\:\mathrm{calculation}\:\mathrm{error}: \\ $$$$\mathrm{3}{t}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{8}{t}_{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\:{t}_{\mathrm{2}} =−\mathrm{3}\vee{t}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$

Commented by Tawa11 last updated on 06/Feb/24

I appreciate sir.

$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

Commented by a.lgnaoui last updated on 07/Feb/24

thank you

$$\mathrm{thank}\:\mathrm{you}\: \\ $$

Answered by cortano12 last updated on 06/Feb/24

determinant ()