Question Number 189201 by Shrinava last updated on 13/Mar/23
$$\mathrm{In}\:\:\:\bigtriangleup\mathrm{ABC}\:\:\:\mathrm{holds}: \\ $$$$\sqrt{\mathrm{2}}\:\mathrm{a}\:\mathrm{cos}\:\frac{\mathrm{B}}{\mathrm{2}}\:\mathrm{cos}\:\frac{\mathrm{C}}{\mathrm{2}}\:=\:\mathrm{s} \\ $$$$\Rightarrow\:\mathrm{sec}\:\left(\mathrm{2B}\right)\:+\:\mathrm{tan}\:\left(\mathrm{2B}\right)\:=\:\frac{\mathrm{c}\:+\:\mathrm{b}}{\mathrm{c}\:−\:\mathrm{b}} \\ $$
Answered by som(math1967) last updated on 13/Mar/23
![(√2)acos(B/2)cos(C/2)=s (√2)a(√((s(s−b))/(ca)))×(√((s(s−c))/(ab)))=s (√2)a(s/a)×(√(((s−b)(s−c))/(bc)))=s (√2) sin(A/2)=1 ∴sin(A/2)=(1/( (√2)))⇒(A/2)=(π/4) ∴A=(π/2) ∴a^2 =b^2 +c^2 now sec2B+tan2B =(1/(cos2B))+((sin2B)/(cos2B)) =((1+sin2B)/(cos2B)) =(((cosB+sinB)^2 )/(cos^2 B−sin^2 B)) =((cosB+sinB)/(cosB−sinB)) =((cosB+sin((π/2)−C))/(cosB−sin((π/2)−C))) [A=(π/2) ∴B+C=(π/2)] =((cosB+cosC)/(cosB−cosC)) =((((c^2 +a^2 −b^2 )/(2ca)) +((a^2 +b^2 −c^2 )/(2ab)))/(((c^2 +a^2 −b^2 )/(2ca)) −((a^2 +b^2 −c^2 )/(2ab)))) =((((2c^2 )/(2ca))+((2b^2 )/(2ab)))/(((2c^2 )/(2ca))−((2b^2 )/(2ab)))) [a^2 =b^2 +c^2 ] =((c+b)/(c−b))](https://www.tinkutara.com/question/Q189218.png)
$$\sqrt{\mathrm{2}}{acos}\frac{{B}}{\mathrm{2}}{cos}\frac{{C}}{\mathrm{2}}={s} \\ $$$$\sqrt{\mathrm{2}}{a}\sqrt{\frac{{s}\left({s}−{b}\right)}{{ca}}}×\sqrt{\frac{{s}\left({s}−{c}\right)}{{ab}}}={s} \\ $$$$\sqrt{\mathrm{2}}{a}\frac{{s}}{{a}}×\sqrt{\frac{\left({s}−{b}\right)\left({s}−{c}\right)}{{bc}}}={s} \\ $$$$\sqrt{\mathrm{2}}\:{sin}\frac{{A}}{\mathrm{2}}=\mathrm{1} \\ $$$$\therefore{sin}\frac{{A}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\Rightarrow\frac{{A}}{\mathrm{2}}=\frac{\pi}{\mathrm{4}} \\ $$$$\therefore{A}=\frac{\pi}{\mathrm{2}}\:\therefore{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${now}\:{sec}\mathrm{2}{B}+{tan}\mathrm{2}{B} \\ $$$$=\frac{\mathrm{1}}{{cos}\mathrm{2}{B}}+\frac{{sin}\mathrm{2}{B}}{{cos}\mathrm{2}{B}} \\ $$$$=\frac{\mathrm{1}+{sin}\mathrm{2}{B}}{{cos}\mathrm{2}{B}} \\ $$$$=\frac{\left({cosB}+{sinB}\right)^{\mathrm{2}} }{{cos}^{\mathrm{2}} {B}−{sin}^{\mathrm{2}} {B}} \\ $$$$=\frac{{cosB}+{sinB}}{{cosB}−{sinB}} \\ $$$$=\frac{{cosB}+{sin}\left(\frac{\pi}{\mathrm{2}}−{C}\right)}{{cosB}−{sin}\left(\frac{\pi}{\mathrm{2}}−{C}\right)}\:\:\left[{A}=\frac{\pi}{\mathrm{2}}\:\therefore{B}+{C}=\frac{\pi}{\mathrm{2}}\right] \\ $$$$=\frac{{cosB}+{cosC}}{{cosB}−{cosC}} \\ $$$$=\frac{\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\:+\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}}{\frac{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{ca}}\:−\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{ab}}} \\ $$$$=\frac{\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{2}{ca}}+\frac{\mathrm{2}{b}^{\mathrm{2}} }{\mathrm{2}{ab}}}{\frac{\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{2}{ca}}−\frac{\mathrm{2}{b}^{\mathrm{2}} }{\mathrm{2}{ab}}}\:\:\:\left[{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right] \\ $$$$=\frac{{c}+{b}}{{c}−{b}} \\ $$
Commented by Shrinava last updated on 13/Mar/23
$$\mathrm{thankyou}\:\mathrm{dearSer}\:\mathrm{cool} \\ $$