ABC triangle′s A+B=C=90° prove that sin^2 A−sin^2 B+sin^2 C=0


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Question Number 131785 by deleteduser12 last updated on 08/Feb/21

$A B C {t r i a n g l e}^{'} s A + B = C = 90 ° p r o v e t h a t$ $\sin^{2} A - {s i n}^{2} B + {s i n}^{2} C = 0$
	Answered by Dwaipayan Shikari last updated on 08/Feb/21

	$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = ϕ$ ${s i n}^{2} A - {s i n}^{2} B + {s i n}^{2} C = \frac{a^{2} + c^{2} - b^{2}}{ϕ^{2}} = Φ$ $A + B = C = \frac{π}{2} \Rightarrow a^{2} + c^{2} = b^{2}$ $Φ = \frac{b^{2} - b^{2}}{ϕ^{2}} = 0$
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