Question Number 131785 by faysal last updated on 08/Feb/21
![ABC triangle′s A+B=C=90° prove that sin^2 A−sin^2 B+sin^2 C=0](https://www.tinkutara.com/question/Q131785.png)
$${ABC}\:{triangle}'{s}\:{A}+{B}={C}=\mathrm{90}°\:{prove}\:{that} \\ $$$$\mathrm{sin}\:^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {B}+{sin}^{\mathrm{2}} {C}=\mathrm{0} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Feb/21
![(a/(sinA))=(b/(sinB))=(c/(sinC))=φ sin^2 A−sin^2 B+sin^2 C=((a^2 +c^2 −b^2 )/φ^2 )=Φ A+B=C=(π/2) ⇒a^2 +c^2 =b^2 Φ=((b^2 −b^2 )/φ^2 )=0](https://www.tinkutara.com/question/Q131786.png)
$$\frac{{a}}{{sinA}}=\frac{{b}}{{sinB}}=\frac{{c}}{{sinC}}=\phi \\ $$$${sin}^{\mathrm{2}} {A}−{sin}^{\mathrm{2}} {B}+{sin}^{\mathrm{2}} {C}=\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\phi^{\mathrm{2}} }=\Phi \\ $$$${A}+{B}={C}=\frac{\pi}{\mathrm{2}}\:\:\:\:\Rightarrow{a}^{\mathrm{2}} +{c}^{\mathrm{2}} ={b}^{\mathrm{2}} \\ $$$$\Phi=\frac{{b}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\phi^{\mathrm{2}} }=\mathrm{0} \\ $$