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Question Number 210375 by Spillover last updated on 08/Aug/24

Commented by som(math1967) last updated on 08/Aug/24

if sin^(−1) x+sin^(−1) y+sin^(−1) z=𝛑 then x^4 +y^4 +z^4 +4x^2 y^2 z^2 =2(x^2 y^2 +y^2 z^2 +z^2 x^2 )

$$\:\boldsymbol{{if}}\:\mathrm{sin}^{−\mathrm{1}} \boldsymbol{{x}}+\mathrm{sin}^{−\mathrm{1}} \boldsymbol{{y}}+\mathrm{sin}^{−\mathrm{1}} \boldsymbol{{z}}=\boldsymbol{\pi} \\ $$$${then}\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} \boldsymbol{{z}}^{\mathrm{2}} \\ $$$$\:=\mathrm{2}\left(\boldsymbol{{x}}^{\mathrm{2}} \boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} \boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \boldsymbol{{x}}^{\mathrm{2}} \right) \\ $$

Commented by som(math1967) last updated on 08/Aug/24

((sinxsiny)/(cosx+cosy))=((2tan(x/2)tan(y/2))/(1−tan^2 (x/2)tan^2 (y/2)))

$$\:\frac{{sinxsiny}}{{cosx}+{cosy}}=\frac{\mathrm{2}{tan}\frac{{x}}{\mathrm{2}}{tan}\frac{{y}}{\mathrm{2}}}{\mathrm{1}−{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{tan}^{\mathrm{2}} \frac{{y}}{\mathrm{2}}} \\ $$

Commented by Spillover last updated on 08/Aug/24

correct.thank you

$${correct}.{thank}\:{you} \\ $$

Commented by Spillover last updated on 08/Aug/24

correct.thank you

$${correct}.{thank}\:{you} \\ $$

Answered by som(math1967) last updated on 08/Aug/24

(c) tan^2 𝛉=1−m^2 ⇒1+tan^2 θ=2−m^2 ⇒sec^2 θ=2−m^2 ⇒secθ=±(2−m^2 )^(1/2) secθ+tan^3 θcosecθ =secθ+((sin^3 θcosecθ)/(cos^3 θ)) =secθ+tan^2 θsecθ =secθ(1+tan^2 θ) =±(2−m^2 )^(1/2) (2−m^2 ) =±(2−m^2 )^(3/2)

$$\left(\boldsymbol{{c}}\right)\:\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{\theta}=\mathrm{1}−{m}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{1}+{tan}^{\mathrm{2}} \theta=\mathrm{2}−{m}^{\mathrm{2}} \\ $$$$\Rightarrow{sec}^{\mathrm{2}} \theta=\mathrm{2}−{m}^{\mathrm{2}} \\ $$$$\Rightarrow{sec}\theta=\pm\left(\mathrm{2}−{m}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:{sec}\theta+{tan}^{\mathrm{3}} \theta{cosec}\theta \\ $$$$={sec}\theta+\frac{{sin}^{\mathrm{3}} \theta{cosec}\theta}{{cos}^{\mathrm{3}} \theta} \\ $$$$={sec}\theta+{tan}^{\mathrm{2}} \theta{sec}\theta \\ $$$$={sec}\theta\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right) \\ $$$$=\pm\left(\mathrm{2}−{m}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{2}−{m}^{\mathrm{2}} \right) \\ $$$$=\pm\left(\mathrm{2}−{m}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$

Commented by Spillover last updated on 08/Aug/24

great.thanks

$${great}.{thanks} \\ $$

Answered by som(math1967) last updated on 08/Aug/24

(e) cosA=((b^2 +c^2 −a^2 )/(2bc)) ⇒2bccosA=b^2 +c^2 −a^2 ⇒2bc(1−2sin^2 (A/2))=b^2 +c^2 −a^2 ⇒a^2 =b^2 +c^2 −2bc+4bcsin^2 (A/2) ∴ a^2 =(b−c)^2 +4bcsin^2 (A/2)

$$\left(\boldsymbol{{e}}\right)\:{cosA}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\Rightarrow\mathrm{2}{bccosA}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2}{bc}\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}}\right)={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}+\mathrm{4}{bcsin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$$$\therefore\:{a}^{\mathrm{2}} =\left({b}−{c}\right)^{\mathrm{2}} +\mathrm{4}{bcsin}^{\mathrm{2}} \frac{{A}}{\mathrm{2}} \\ $$

Commented by Spillover last updated on 08/Aug/24

great.thanks

$${great}.{thanks} \\ $$

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