Question-173002

Question Number 173002 by Mikenice last updated on 04/Jul/22

Commented by kaivan.ahmadi last updated on 05/Jul/22

1−tg^2 θ=1−((sin^2 θ)/(cos^2 θ))=((cos^2 θ−sin^2 θ)/(cos^2 θ))= ((1−2sin^2 θ)/(1−sin^2 θ))=((1−2(((a−b)/(a+b)))^2 )/(1−(((a−b)/(a+b)))^2 ))=(((a+b)^2 −2(a−b)^2 )/((a+b)^2 −(a−b)^2 ))= ((−a^2 +6ab−b^2 )/(4ab))

$$\mathrm{1}−{tg}^{\mathrm{2}} \theta=\mathrm{1}−\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}=\frac{{cos}^{\mathrm{2}} \theta−{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{2}} \theta}= \\ $$$$\frac{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \theta}{\mathrm{1}−{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{1}−\mathrm{2}\left(\frac{{a}−{b}}{{a}+{b}}\right)^{\mathrm{2}} }{\mathrm{1}−\left(\frac{{a}−{b}}{{a}+{b}}\right)^{\mathrm{2}} }=\frac{\left({a}+{b}\right)^{\mathrm{2}} −\mathrm{2}\left({a}−{b}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }= \\ $$$$\frac{−{a}^{\mathrm{2}} +\mathrm{6}{ab}−{b}^{\mathrm{2}} }{\mathrm{4}{ab}} \\ $$

Answered by CElcedricjunior last updated on 05/Jul/22

sin𝛉=((a−b)/(a+b)) calculons 1−tan^2 𝛉 1−tan^2 𝛉=1−((sin^2 𝛉)/( 1−sin^2 𝛉)) =1−(((a−b)^2 )/((a+b)^2 ))×(((a+b)^2 )/(4ab)) =((−(a^2 −6ab+b^2 ))/(4ab)) 1−tan^2 𝛉=((−(a−3b−2b(√(2 )))(a−3b+2b(√2)))/(4ab)) .........Le ce^� le^� bre cedric junior............

$$\boldsymbol{\mathrm{sin}\theta}=\frac{\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}}{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}} \\ $$$$\boldsymbol{\mathrm{calculons}}\:\mathrm{1}−\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\mathrm{1}−\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\theta}=\mathrm{1}−\frac{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\theta}}{\:\mathrm{1}−\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{\theta}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1}−\frac{\left(\boldsymbol{\mathrm{a}}−\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} }×\frac{\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)^{\mathrm{2}} }{\mathrm{4}\boldsymbol{\mathrm{ab}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} −\mathrm{6}\boldsymbol{\mathrm{ab}}+\boldsymbol{\mathrm{b}}^{\mathrm{2}} \right)}{\mathrm{4}\boldsymbol{\mathrm{ab}}} \\ $$$$\:\mathrm{1}−\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\theta}=\frac{−\left(\boldsymbol{\mathrm{a}}−\mathrm{3}\boldsymbol{\mathrm{b}}−\mathrm{2}\boldsymbol{\mathrm{b}}\sqrt{\mathrm{2}\:\:}\right)\left(\boldsymbol{\mathrm{a}}−\mathrm{3}\boldsymbol{\mathrm{b}}+\mathrm{2}\boldsymbol{\mathrm{b}}\sqrt{\mathrm{2}}\right)}{\mathrm{4}\boldsymbol{\mathrm{ab}}} \\ $$$$ \\ $$$$ \\ $$$$………\mathscr{L}{e}\:{c}\acute {{e}l}\grave {{e}bre}\:{cedric}\:{junior}………… \\ $$

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